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How to algebraically find the intersection of two lines

How to algebraically find the intersection of two lines

An intersection of two lines is a point where the graphs of two lines cross each other. Every pair of lines does have an intersection, except if the lines are parallel. This means that the lines move in the same direction. You can check whether two lines are parallel by determining their slope. If the slopes are equal, then the lines are parallel. This means they do not cross each other, or if the lines are the same then they cross in every point. You can determine the slope of a line with the help of the derivative.

Every line can be represented with the expression y = ax + b, where x and y are the two-dimensional coordinates and a and b are constants that characterize this specific line.

For a point (x,y) to be an intersection point we must have that (x,y) lays on both lines, or in other words: If we fill in these x and y than y = ax + b must be true for both lines.

An Example of Finding the Intersection of Two Lines

Let's look at two lines:

Then we must find a point (x,y) that satisfies both linear expressions. To find such a point we must solve the linear equation:

To do this, we must write the variable x to one side, and all terms without x to the other side. So the first step is to subtract 4x on both sides of the equality sign. Since we subtract the same number on both the right hand side as well as the left hand side the solution does not change. We get:

3x + 2 – 4x = 4x – 9 -4x

Then we subtract 2 on both sides to get:

Finally, we multiply both sides with -1. Again, since we perform the same operation on both sides the solution does not change. We conclude x = 11.

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We had y = 3x + 2 and fill in x = 11. We get y = 3*11 + 2 = 35. So the intersection is at (7,11). If we check the second expression y = 4x – 9 = 4*11 -9 = 35. So indeed we see that the point (7,11) also lies on the second line.

In the picture below, the intersection is visualized.

How to algebraically find the intersection of two lines

Parallel Lines

To illustrate what happens if the two lines are parallel there is the following example. Again we have two lines, but this time with the same slope.

Now if we want to solve 2x + 5 = 2x + 3 we have a problem. It is impossible to write all terms involving x to one side of the equality sign since we then would have to subtract 2x from both sides. However if we would do this we end up with 5 = 3, which clearly is not true. Therefore this linear equation has no solution and hence there is no intersection between these two lines.

Other Intersections

Intersections do not limit to two lines. We can calculate the intersection point between all types of curves. If we look further than only lines we might get situations in which there are more than one intersection. There are even examples of combinations of functions that have infinitely many intersections. For example the line y = 1 (so y = ax + b where a = 0 and b = 2) has infinitely many intersections with y = cos(x) since this function oscillates between -1 and 1.

Here, we will look at an example of the intersection between a line and a parabola. A parabola is a curve which is represented by the expression y = ax 2 + bx + c. The method of finding the intersection remains roughly the same. Let's for example look at the intersection between the following two curves:

Again we equate the two expressions and we look at 3x + 2 = x 2 + 7x – 4.

We rewrite this to a quadratic equation such that one side of the equality sign is equal to zero. Then we must find the roots of the quadratic function we get.

So we start by subtracting 3x + 2 on both sides of the equality sign:

There are multiple ways to find the solutions of these kind of equations. If you want to know more about these solution methods I suggest reading my article about finding the roots of a quadratic function. Here we will choose to complete the square. In the article about quadratic functions I describe in detail how this method works, here we will just apply it.

Then the solutions are x = -2 + sqrt 10 and x = -2 – sqrt 10.

Now we will fill in this solution in both expressions to check whether this is correct.

y = 3*(-2 + sqrt 10) + 2 = – 4 + 3 * sqrt 10

y = (-2 + sqrt 10) 2 + 7*(-2 + sqrt 10) – 4 = 14 – 4*sqrt 10 -14 + 7*sqrt 20 – 4

So indeed, this point was an intersection point. One can also check the other point. This will result in the point (-2 – sqrt 10, -4 – 3*sqrt 10). It is important to make sure you check the right combinations if there are multiple solutions.

It always helps to draw the two curves to see if what you calculated makes sense. In the picture below you see the two intersection points.

How to algebraically find the intersection of two lines

Summary

To find the intersection between two lines y = ax + b and y = cx + d the first step that must be done is to set ax + b equal to cx + d. Then solve this equation for x. This will be the x coordinate of the intersection point. Then you can find the y coordinate of the intersection by filling in the x coordinate in the expression of either of the two lines. Since it is an intersection point both will give the same y coordinate.

It is also possible to calculate the intersection between other functions, which are not lines. In these cases it might happen that there is more than one intersection. The method of solving remains the same: set both expression equal to each other and solve for x. Then determine y by filling in x in one of the expressions.

This content is accurate and true to the best of the author’s knowledge and is not meant to substitute for formal and individualized advice from a qualified professional.

How to algebraically find the intersection of two lines

In mathematics, we refer to the point of intersection where a point meets two lines or curves. The intersection of lines may be an empty set, a point, or a line in Euclidean geometry . A required criterion for two lines to intersect is that they should be in the same plane and are not skew lines. The intersection formula gives the point at which these lines are meeting.

Point of Intersection of Two Lines Formula

Consider 2 straight lines \[a_<1>x+b_<1>y+c_<1>\] and \[a_<2>x+b_<2>y+c_<2>\] which are intersecting at point (x,y) as shown in figure. So, we have to find a line intersection formula to find these points of intersection (x,y). These points satisfy both the equations.

By solving these two equations we can find the intersection of two lines formula.

The formula for the point of intersection of two lines will be as follows:

So, this is the point of intersection formula of 2 lines when intersecting at one point.

A Intersection B Intersection C Formula

If we have 3 sets A,B and C. A intersection B intersection C formula will be as follows:

\[n(A \cup B \cap C)=n(A) + n(B) + n(C) − n(A \cap B) − n(B \cap C) − n(C \cap A) + n(A \cap B \cap C)\]

The Formula of A Intersection B

If we have 2 sets, say A and B. The formula of A intersection B will be as follows:

\[n(A \cup B)=n(A) + n(B) − n(A \cap B) \]

\[n(A \cup B)\] is the number of elements present in either set A or set B.

\[n(A \cap B)\] is the number of elements present in both set A and set B.

Problems on Point of Intersection of Two Lines Formula:

1. Find the Point of Intersection of Two Lines \[2x + 4y + 6 = 0\] and \[2x + 3y + 4 = 0\].

Ans: The two line equations are given as \[2x + 4y + 6 = 0\] and \[2x + 3y + 4 = 0\].

Comparing these two equations with \[a_<1>x+b_<1>y+c_ <1>and a_<2>x+b_<2>y+c_<2>\]

Point of intersection formula is given as

Substituting the values of \[a_<1>,b_<1>,c_<1>,a_<2>,b_<2>,c_<2>\] in the intersection formula

Therefore the point of intersection of two lines \[2x + 4y + 6 = 0\] and \[2x + 3y + 4 = 0\] is \[(x,y)=(1,−2)\].

The point of intersection formula is used to find the meeting point of two lines, also known as the point of intersection. The equation can be used to represent these two lines \[a_<1>x + b_<1>y + c_ <1>= 0\] and \[a_<2>x + b_<2>y + c_ <2>= 0\] respectively. The point of intersection of three or more lines can be found. We can discover the solution for the point of intersection of two lines by solving the two equations. Let’s look at some solved examples of the point of intersection formula.

Two straight lines will intersect at a point if they are not parallel. The point of intersection is the meeting point of two straight lines.

If two crossing straight lines have the same equations, the intersection point can be found by solving both equations at the same time.

On a two-dimensional graph, straight lines intersect at just one location, which is described by a single set of display style x-coordinates and display style y-coordinates. You know the display style x-coordinates and display style y-coordinates must fulfil both equations since both lines pass through that location.

Intersection Point

Have you ever been driving and come across a traffic sign like this?

This is an intersection traffic sign, and it means you’re approaching a place where two roads connect. Two lines cross and meet in the centre of the intersection traffic sign, as you can see. This is where their paths cross. The intersection of two lines or curves is referred to as a point of intersection in mathematics.

The intersection of two curves is noteworthy because it is the point at which the two curves have the same value. This can come in handy in a variety of situations. Let’s imagine we’re dealing with an equation that represents a company’s revenue and another that represents the company’s expense. The point of intersection of the curves corresponding to these two equations is where revenue equals cost; this is the company’s breakeven point.

Application for a Map of Points of Intersection

A point of intersection is the meeting point of two lines or curves.

By graphing the curves on the same graph and finding their points of intersection, we can discover a point of junction graphically.

The following steps can be used to find a point of intersection algebraically:

For one of the variables, call it y, and solve each equation.

Set the equations for y discovered in the first step to the same value, then solve for the other variable, which we’ll call x. This is the x-value of the junction point.

In any of the original equations, plug in the x-value of the site of intersection and solve for y. This is the point of intersection y-variable.

Example: Find the point of intersection of two lines \[2x + 4y + 2 = 0\] and \[2x + 3y + 5 = 0\] using the point of intersection formula.

Given Straight Line Equations are:

\[2x + 4y + 2 = 0\] and \[2x + 3y + 5 = 0\]

\[a_ <1>= 2, b_ <1>= 4, c_ <1>= 2\]

\[a_ <2>= 2, b_ <2>= 3, c_ <2>= 5\]

Intersection point can be calculated using the point of intersection formula,

Lines that are non-coincident and non-parallel intersect at a unique point. Lines are said to intersect each other if they cut each other at a point. By Euclid’s lemma two lines can have at most 1 1 1 point of intersection. In the figure below lines L 1 L1 L 1 and L 2 L2 L 2 intersect each other at point P . P. P . Three or more lines when met at a single point are said to be concurrent and the point of intersection is point of concurrency.

How to algebraically find the intersection of two lines

Intersection of Lines

Point of Intersection

Find the intersection of the lines y = 3 x − 3 y = 3x – 3 y = 3 x − 3 and y = 2.3 x + 4 y = 2.3x + 4 y = 2 . 3 x + 4 .

Other Properties

Angle between the lines:

For two lines intersecting at right angle, m 1 m 2 = − 1. < m >_< 1 >< m >_ < 2 >=-1. m 1 ​ m 2 ​ = − 1 .

However, general equation in degree 2 2 2 a x 2 + 2 h x y + b y 2 + 2 g x + 2 f y + c = 0 ax^2+2hxy+by^2+2gx+2fy+c=0 a x 2 + 2 h x y + b y 2 + 2 g x + 2 f y + c = 0 will represent a pair of straight lines if and only if a b c + 2 f g h − a f 2 − b g 2 − c h 2 = 0 and h 2 − a b > 0. abc+ 2fgh- af^2 -bg^2 -ch^2 =0\quad \text< and >\quad h^2 – ab > 0. a b c + 2 f g h − a f 2 − b g 2 − c h 2 = 0 and h 2 − a b > 0 . The angle θ \theta θ between these lines satisfies tan ⁡ θ = h 2 − a b ∣ a − b ∣ . \tan \theta=\frac<\sqrt><|a-b|>. tan θ = ∣ a − b ∣ h 2 − a b

Combined Equation of Pair of Lines joining Origin and Intersection Points of a Curve and a Line

Let us find the equation of the straight lines joining the origin and the points of intersection of the curve

a x 2 + 2 h x y + b y 2 + 2 g x + 2 f y + c = 0 ax^2+2hxy+by^2+2gx+2fy+c=0 a x 2 + 2 h x y + b y 2 + 2 g x + 2 f y + c = 0

and proceed as above, solving for x, then substituting that value into either equation to find y.

The two equations need not even be in the same form. Just set them equal to each other and proceed in the usual way.

When one line is vertical

When one of the lines is vertical, it has no defined slope, so its equation will look something like x=12 . See Vertical lines (Coordinate Geometry). We find the intersection slightly differently. Suppose we have the lines whose equations are

y = 3x-3 A line sloping up and to the right
x = 12 A vertical line

On the vertical line, all points on it have an x-coordinate of 12 (the definition of a vertical line), so we simply set x equal to 12 in the first equation and solve it for y.
Equation for a line:

So the intersection point is at (12,33).

If both lines are vertical, they are parallel and have no intersection (see below).

When they are parallel

Segments and rays might not intersect at all

In the case of two non-parallel lines, the intersection will always be on the lines somewhere. But in the case of line segments or rays which have a limited length, they might not actually intersect.

In Fig 1 we see two line segments that do not overlap and so have no point of intersection. However, if you apply the method above to them, you will find the point where they would have intersected if extended enough.

Things to try

  1. In the above diagram, press ‘reset’.
  2. Drag any of the points A,B,C,D around and note the location of the intersection of the lines.
  3. Drag a point to get two parallel lines and note that they have no intersection.
  4. Click ‘hide details’ and ‘show coordinates’. Move the points to any new location where the intersection is still visible. Calculate the slopes of the lines and the point of intersection. Click ‘show details’ to verify your result.

Limitations

In the interest of clarity in the applet above, the coordinates are rounded off to integers and the lengths rounded to one decimal place. This can cause calculatioons to be slightly off.

Equate right hand sides (with one squaring) solve numerically.

$$ x \, e^ <6 x >= 1 \Rightarrow x \approx 0.2387 $$

How to algebraically find the intersection of two lines

The graph below shows it is not some simple solution:

How to algebraically find the intersection of two lines

The tiny font says $A=(0.24,0.49)$ with the rounded coordinates of the intersection point.

Building on Narasimham’s answer, one might try the Lambert W function here, on $$ 6 = 6x \, e^ <6x>= f(6x) \Rightarrow \\ W(6) = W(f(6x)) = 6x \Rightarrow \\ x = W(6)/6 $$

How to algebraically find the intersection of two lines

Solve $$\sqrt=e^<-3x>;$$ for $x$. Divide both sides by $e^<-3x>$: $$\frac<\sqrt>>=1;$$ rewrite $\sqrt/e^<-3x>=\sqrte^<3x>$: $$\sqrte^<3x>=1;$$ square both sides: $$xe^<6x>=1;$$ multiply both sides by $6$: $$6xe^<6x>=6;$$ substitute $u=6x$: $$ue^u=6;$$ take the product logarithms of both sides: $$u=\operatorname_k(6);$$ substitute back for $u$: $$6x=\operatorname_k(6);$$ divide both sides by $6$: $$x=\frac<\operatorname_k(6)><6>;$$ write out the answer: $$\therefore x=\boxed<\frac<\operatorname_k(6)><6>>\phantom<.>,$$ where $\operatorname_k$ is the $k$-th branch of the product log function.

The solution is real for $k=0$: $$x\in\mathbb\implies x=\frac<\operatorname_0(6)><6>=\frac<\operatorname(6)><6>\approx0.2387341293163833852.$$

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Two distinct lines intersect at the most at one point. To find the intersection of two lines we need the general form of the two equations, which is written as \(\beginx + y + = 0, \text < and >x + y + = 0\end\). The lines will intersect only if they are non-parallel lines. Common examples of intersecting lines in real life include a pair of scissors, a folding chair, a road cross, a signboard, etc. In this mini-lesson, we will learn in detail, how to find the point of intersection of two lines.

1. Meaning of Intersection of Two Lines
2. Finding Intersection of Two Lines
3. The Angle of Intersection
4. Conditions for Two Lines to be Parallel or Perpendicular
5. Properties of Intersecting Lines
6. Solved Examples on Intersection of Two Lines
7. Practice Questions on Intersection of Two Lines
8. FAQS on Intersection of Two Lines

Meaning of Intersection of Two Lines

When two lines share exactly one common point, they are called the intersecting lines. The intersecting lines share a common point. And, this common point that exists on all intersecting lines is called the point of intersection. The two non-parallel straight lines which are co-planar will have an intersection point. Here, lines A and B intersect at point O, which is the point of intersection.

How to algebraically find the intersection of two lines

Finding Intersection of Two Lines

Let’s consider the following case. We are given two lines, \(\) and \(\), and we are required to find the point of intersection. Evaluating the point of intersection involves solving two simultaneous linear equations.

Let the equations of the two lines be (written in the general form): \(\beginx + y + = 0\\x + y + = 0\end\)

How to algebraically find the intersection of two lines

Now, let the point of intersection be \(\left( <,> \right)\). Thus,

This system can be solved using Cramer’s rule to get:

From this relation, we can obtain the point of intersection \(\left( <,> \right)\) as

The Angle of Intersection

To obtain the angle of intersection between two lines, consider the figure shown:

The equations of the two lines in slope-intercept form are:

Note in the figure above that \(\theta = <\theta _2>– <\theta _1>\), and thus

Conventionally, we would be interested only in the acute angle between the two lines and thus, we have to have \(\tan \theta \) as a positive quantity.

So in the expression above, if the expression \(\frac <<>><<1 + >>\) turns out to be negative, this would be the tangent of the obtuse angle between the two lines; thus, to get the acute angle between the two lines, we use the magnitude of this expression.

Therefore, the acute angle \(\theta \) between the two lines is

From this relation, we can easily deduce the conditions on \(\) and \(\) such that the two lines \(\) and \(\) are parallel or perpendicular.

Conditions for Two Lines to be Parallel or Perpendicular

If the lines are parallel, \(\theta = 0\) and \( = \), which is obvious since parallel lines must have the same slope.

For the two lines to be perpendicular lines, θ = π/2 , so that cot θ = 0; this can happen if \(1 + = 0\) or \( = – 1\).

If the lines \(\) and \(\) are in the general form ax + by + c = 0, the slope of this line is m = -a/b.

Condition for Two Lines to be Parallel

Thus, the condition for \(\) and \(\) to be parallel is:

Example

The line \(😡 – 2y + 1 = 0\) is parallel to the line \(😡 – 2y – 3 = 0\) because the slope of both the lines is m = 1/2

Condition for Two Lines to be Perpendicular

The condition for \(\) and \(\) to be perpendicular is:

Example

The line \(\) : x + y = 1 is perpendicular to the line \(\) 😡 – y = 1 because the slope of \(\) is \( – 1\) while the slope of \(\) is 1.

Properties of Intersecting Lines

  • The intersecting lines (two or more) always meet at a single point.
  • The intersecting lines can cross each other at any angle. This angle formed is always greater than 0 ∘ and less than 180 ∘ .
  • Two intersecting lines form a pair of vertical angles. The vertical angles are opposite angles with a common vertex (which is the point of intersection).

How to algebraically find the intersection of two lines

Here, ∠a and ∠c are vertical angles and are equal.

Also, ∠b and ∠d are vertical angles and equal to each other.

∠a+∠d = straight angle = 180 ∘

An acute angle \(\theta \) between lines \(L_1\) and \(L_2\) with slopes \(m_1\) and \(m_2\) is given by

If the lines \(\) and \(\) are given in the general form ax + by + c = 0, the slope of this line is m = -a/b.

The condition for two lines \(\) and \(\) to be parallel is:

The condition for two lines \(\) and \(\) to be perpendicular is:

Related articles on Intersection of Two Lines

Check out the articles below to know more about topics related to the intersection of two lines.

Solved Examples on Intersection of Two Lines

Example 1: Find the point of intersection and the angle of intersection for the following two lines:

How to algebraically find the intersection of two lines

x – 2y + 3 = 0
3x – 4y + 5 = 0

Solution:

We use Cramer’s rule to find the point of intersection:

⇒ x = 1, y = 2
Now, the slopes of the two lines are:

If \(\theta \) is the acute angle of intersection between the two lines, we have:

Point of intersection is (1,2).
The angle of intersection is θ = tan −1 (2/11)

Example 2: Find the equation of a line perpendicular to the line x – 2y + 3 = 0 and passing through the point (1, -2).

How to algebraically find the intersection of two lines

Solution:

Given line x – 2y + 3 = 0 can be written as

Slope of the line 1 is \(\)= 1/2

Therefore, slope of the line perpendicular to line \((1)\) is

Equation of a line perpendicular to the line x – 2y + 3 = 0 and passing through the point (1, -2) is

People also ask, what is the point of intersection called?

Point of Intersection. Definition : Two lines intersect when they cross each other. They form vertically opposite angles, which we will learn later. The point where the lines intersect is called the point of intersection. If the angles produced are all right angles, the lines are called perpendicular lines.

  1. Get the two equations for the lines into slope-intercept form.
  2. Set the two equations for y equal to each other.
  3. Solve for x.
  4. Use this x-coordinate and plug it into either of the original equations for the lines and solve for y.

Similarly, how do you solve system of equations?

  1. Step 1: Multiply the entire first equation by 2.
  2. Step 2: Rewrite the system of equations, replacing the first equation with the new equation.
  3. Step 3: Add the equations.
  4. Step 4: Solve for x.
  5. Step 5: Find the y-value by substituting in 3 for x in either equation.

How do you find the Y intercept?

To find the y intercept using the equation of the line, plug in 0 for the x variable and solve for y. If the equation is written in the slope-intercept form, plug in the slope and the x and y coordinates for a point on the line to solve for y.

Here we will cover a method for finding the point of intersection for two linear functions . That is, we will find the (x, y) coordinate pair for the point were two lines cross.

Our example will use these two functions:

We will call the first one Line 1, and the second Line 2. Since we will be graphing these functions on the x, y coordinate axes, we can express the lines this way:

Note that these two lines are in slope-intercept form.

These two lines look this way:

Now, where the two lines cross is called their point of intersection. Certainly this point has (x, y) coordinates. It is the same point for Line 1 and for Line 2. So, at the point of intersection the (x, y) coordinates for Line 1 equal the (x, y) coordinates for Line 2.

Since at the point of intersection the two y-coordinates are equal, (we’ll get to the x-coordinates in a moment), let’s set the y-coordinate from Line 1 equal to the y-coordinate from Line 2.

The y-coordinate for Line 1 is calculated this way:

The y-coordinate for Line 2 is calculated this way:

Setting the two y-coordinates equal looks like this:

Now, we do some algebra to find the x-coordinate at the point of intersection:

2x + 3 = -0.5x + 7 We start here.
2.5x + 3 = 7 Add 0.5x to each side.
2.5x = 4 Subtract 3 from each side.
x = 4/2.5 Divide each side by 2.5.
x = 1.6 Divide 4 by 2.5.

So, we have the x-coordinate for the point of intersection. It’s x = 1.6. Now, let’s find the y-coordinate. The y-coordinate can be found by placing the x-coordinate, 1.6, into either of the equations for the lines and solving for y. We will first use the equation for Line 1:

Therefore, the y-coordinate is 6.2. To make sure our calculations are correct, and also to demonstrate a point, we should get the same y-coordinate if we use the equation for Line 2. Let’s try that:

Well, looks like everything has worked out. The point of intersection for these two lines is (1.6, 6.2). If you look back at the graph, this certainly makes sense:

Here’s the summary of our methods:

  1. Get the two equations for the lines into slope-intercept form. That is, have them in this form: y = mx + b.
  2. Set the two equations for y equal to each other.
  3. Solve for x. This will be the x-coordinate for the point of intersection.
  4. Use this x-coordinate and plug it into either of the original equations for the lines and solve for y. This will be the y-coordinate of the point of intersection.
  5. As a check for your work plug the x-coordinate into the other equation and you should get the same y-coordinate.
  6. You now have the x-coordinate and y-coordinate for the point of intersection.

Actually, there is nothing special about the functions being linear functions. This method could be used to find the point or points of intersection between many other types of functions. One would express the functions in ‘y =’ form, set the right side of these forms equal to each other, solve for x, (or x’s), and use this x, (or x’s), to find the corresponding y, (or y’s).

Here’s a calculator to help you check your work. Make up two linear functions in slope-intercept form. Calculate the point of intersection using the above methods. Then enter the slope and y-intercept for each line into the calculator and click the button to check your work.

When two lines or more than two lines cross each other in a plane, this is called the intersection of lines, and the lines are called intersecting lines. They have a common point called the point of intersection.

What is the Point of Intersection Calculator?

‘Cuemath’s Point of Intersection Calculator’ is an online tool that helps to calculate the value of intersection points for the given equations. Cuemath’s online Point of Intersection Calculator helps you to calculate the value of intersection points for the given equations in a few seconds.

How to Use Point of Intersection Calculator?

Please follow the steps below on how to use the calculator:

  • Step 1: Enter the coefficients of equations in the given input box.
  • Step 2: Click on the “Calculate” button to find the value of the intersection point.
  • Step 3: Click on the “Reset” button to clear the fields and enter new values.

How to Find Point of Intersection?

When two lines share exactly one common point, they are called intersecting lines. The intersecting lines share a common point. And, this common point that exists on all intersecting lines is called the point of intersection.

An equation of the form Ax + By = C. Here, x and y are variables, and A, B, and C are constants.

To solve the point of intersection of the given equations, let’s see an example to understand briefly

How to algebraically find the intersection of two lines

Solved Example:

Solve 2x + y = 7 and x + y = 5

Solution:

From (2), x = 5 – y ——> (3)

Substitute (3) in (1),

Substitute (4) in (2),

Therefore, point of intersection (x, y) = (2, 3)

Similarly, you can try the calculator to find the value of intersection point for the given equations:

So I have several problems that ask me to find all points of intersection algebraically, but I haven’t been able to make much headway on most of them.

The first problem

Homework Statement

Find all the points of intersection algebraically of the graphs of . on the interval [0, 4π]

Homework Equations

f(x) = sinx + 1
g(x) = cosx

The Attempt at a Solution

-Attempt #1:
sinx + 1 = cosx
sinx + 1 – cosx = 0
sinx – cosx = -1

-Attempt #2:
sinx + 1 = cosx
sinx / cosx + 1 / cosx = cosx / cosx
tanx + secx = 1
sinx/cosx + 1/cosx = 1

-Attempt #3:
sinx + 1 = cosx
sinx^2 + 1 = cosx^2
sinx^2 + 1 = 1 – sinx^2
2sinx^2 + 1 = 1
2sinx^2 = 0

. I think I’m really close to the answer here, but I’m not sure where to go now.

The second problem

Homework Statement

Find all the points of intersection algebraically of the graphs of . on the interval [0, 4π]

Homework Equations

f(x) = tanx
g(x) = sinx

The Attempt at a Solution

sinx = tanx
sinx = sinx / cosx
sinx * cosx – sinx = 0
sinx (cosx – 1) = 0

sinx = 0
cosx = 1

It’s (1,0) at [0, 2π, 4π], but apparently [π, 3π] are also points of intersection, even though the cosine value at [π, 3π] are -1. Could somebody explain that please? (I’ve pretty much already solved it)

The third problem

Homework Statement

Find all the points of intersection algebraically of the graphs of . on the interval [0, 4π]

Homework Equations

f(x) = tanx
g(x) = cotx

The Attempt at a Solution

Attempt #1:
tanx = cotx
sinx / cosx = cosx / sinx
sinx^2 / cosx = cosx
sinx^2 = cosx^2
sinx^2 = -sinx^2 + 1
2sinx^2 = 1

And again . I think I’m really close to the answer here, but I’m not sure where to go now.

Any helpful advice or tips would be really appreciated! Thankyou.

Answers and Replies

  • Jul 15, 2014
  • #2

Question 1:
You used the correct method to obtain the result (which is to square both sides and then substitute ##\cos^2x## for ##1-\sin^2x##), however you made a mistake when simplifying since ##(\sin x+1)^2\neq \sin^2 x+1##. To obtain the correct result you may use the following identity: ##(a+b)^2=a^2+2ab+b^2##.

Question 2:
##\sin x=0\,\text< or >\,\cos x=1## does not mean that ##y = 0, x = 1##, can you see where you made the mistake?

Question 3:
Let me help you again: ##2\sin^2x = 1\Rightarrow 2\sin^2x-1=0\Rightarrow\left(\sqrt<2>\sin x-1\right)\left(\sqrt<2>\sin x+1\right)=0.## Do you know how to proceed?

  • Jul 15, 2014
  • #3
  • Jul 15, 2014
  • #4
  • Jul 15, 2014
  • #5

Mmm, I think I’m still a little slow on the uptake. I still don’t really understand why [π, 3π] work, because the cos values of [π, 3π] are both -1, and if cosx = 1, then that doesn’t make the equation true, since -1 doesn’t equal 1. I think I’m really missing some fundamental basic idea here though .

cosx = 1
cos(pi) = -1
cos(3pi) = -1

  • Jul 15, 2014
  • #6

Do you know how to solve equations of the form ##a\cos x+b=0## or ##a\sin x+b=0##? If not then the following may help you a bit:
##a\cos x=b\Rightarrow \cos x=\tfrac ba.## What this equation equation means is that we’re looking for a number ##x## such that ##\cos x## is equal to ##b/a##. To find this number we can use our classical trigonometric values table:

So for example if we want to solve the equation ##2\cos x=\sqrt<2>##, we proceed as follows:
##2\cos x=\sqrt<2>\Rightarrow \cos x=\tfrac<\sqrt<2>>2.## We look at our trigonometric table and we find that the angle whose cosine corresponds to ##\tfrac<\sqrt<2>>2## is ##\tfrac\pi4##. So ##\tfrac\pi4## solves our equation. But it’s not the only solution, since numbers of the form ##\tfrac\pi4+2n\pi## (where ##n## is an integer) are also solutions.

If it is not in our table then we use the ##\arccos## function. For example if you have an equation like ##3\cos x=1\Rightarrow \cos x=\tfrac13\,(1)##, you can see that ##\tfrac13## isn’t in our table. To solve this problem we use the ##\arccos## function which is like the inverse of ##\cos##, more precisely ##\arccos(\cos x)=x##. So by ##(1)## we have that ##\arccos(\cos x)=\arccos \tfrac13\Rightarrow x=\arccos \tfrac13##. And remember not only ##\arccos \tfrac13## is a solution but also numbers of the form ##\arccos \tfrac13+2n\pi##. (where ##n## is an integer)

Estimating where two functions intersect using data

Posted July 04, 2013 at 02:38 PM | categories: data analysis | tags:

Updated July 07, 2013 at 09:01 AM

Suppose we have two functions described by this data:

T(K) E1 E2
300 -208 -218
400 -212 -221
500 -215 -220
600 -218 -222
700 -220 -222
800 -223 -224
900 -227 -225
1000 -229 -227
1100 -233 -228
1200 -235 -227
1300 -240 -229

We want to determine the temperature at which they intersect, and more importantly what the uncertainty on the intersection is. There is noise in the data, which means there is uncertainty in any function that could be fit to it, and that uncertainty would propagate to the intersection. Let us examine the data.

How to algebraically find the intersection of two lines

Our strategy is going to be to fit functions to each data set, and get the confidence intervals on the parameters of the fit. Then, we will solve the equations to find where they are equal to each other and propagate the uncertainties in the parameters to the answer.

These functions look approximately linear, so we will fit lines to each function. We use the regress function in pycse to get the uncertainties on the fits. Then, we use the uncertainties package to propagate the uncertainties in the analytical solution to the intersection of two lines.

How to algebraically find the intersection of two lines

You can see there is a substantial uncertainty in the temperature at approximately the 90% confidence level (± 2 σ).

After a suggestion from Prateek, here we subtract the two data sets, fit a line to that data, and then use fsolve to find the zero. We wrap fsolve in the uncertainties package to directly get the uncertainty on the root.

Interesting that this uncertainty is a little smaller than the previously computed uncertainty. Here you can see we have to wrap the function in a peculiar way. The function must return a single float number, and take arguments with uncertainty. We define the polynomial fit (a line in this case) in a lambda function inside the function. It works ok.

Algebraically determine the point(s) of intersection between the following functions.
f(x) = 2x^2+3
g(x) = −3x+8

  1. ℹ️

2x^2+3 = −3x+8
2x^2 + 3x – 5 = 0
(2x+5)(x-1) = 0
.

  1. ℹ️

Respond to this Question

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A line can cut through a parabola in two points, or it may just be tangent to the parabola and touch it at one point. And then, sadly, a line and a parabola may never meet. When solving systems of equations involving lines and parabolas, you usually use the substitution method — solving for x or y in the equation of the line and substituting into the equation of the parabola.

Sometimes the equations lend themselves to elimination — when adding the equations (or multiples of the equations) together eliminates one of the variables entirely because its coefficient becomes 0. Elimination works only occasionally, but substitution always works.

Sample questions

Find the common solution(s) in the equations y = –5x 2 + 12x + 3 and 8x + y = 18.

The points of intersection are (1, 10), (3, –6). Here’s another way to write this solution: When x = 1, y = 10, and when x = 3, y = –6. To find these solutions, rewrite the equation of the line as y = 18 – 8x.

Replace the y in the equation of the parabola with its equivalent to get 18 – 8x = –5x 2 + 12x + 3. Move all the terms to the left and combine like terms, giving you 5x 2 – 20x + 15 = 0. Divide each term by 5 and then factor, which gives you the equation 5(x 2 – 4x + 3) = 5(x – 3)(x – 1) = 0.

Using the multiplication property of zero (in order for a product to equal 0, one of the factors must be 0), you know that x = 3 or x = 1. Substitute those values back into the equation of the line to get the corresponding y-values.

Always substitute back into the equation with the lower exponents. You can avoid creating extraneous solutions.

Find the common solution(s) in the equations y = x 2 – 4x and 2x + y + 1 = 0

(1, –3). Solve for y in the equation of the line to get y = –2x – 1. Substitute this value into the equation of the parabola to get –2x – 1 = x 2 – 4x. Moving the terms to the right and simplifying, 0 = x 2 – 2x + 1 = (x – 1) 2 .

The only solution is x = 1. Replacing x with 1 in the equation of the line, you find that y = –3. The line is tangent to the parabola at the point of intersection, which is why this problem has only one solution.

Practice questions

Find the common solution(s) in the equations y = x 2 + 4x + 7 and 3xy + 9 = 0.

Find the common solution(s) in the equations y = 4x 2 – 8x – 3 and 4x + y = 5.

Following are answers to the practice questions:

The answer is (–2, 3), (1, 12).

Solve for y in the second equation (you get y = 3x + 9), and substitute that into the equation of the parabola: 3x + 9 = x 2 + 4x + 7. Move all the terms to the right and factor the equation: 0 = x 2 + x – 2 = (x + 2)(x – 1).

So, x = –2 or 1. Letting x = –2 in the equation of the line, 3(–2) – y + 9 = 0; –6 – y + 9 = 0; –y = –3; y = 3. And when x = 1 in the equation of the line, 3(1) – y + 9 = 0; 3 – y + 9 = 0; –y = –12; y = 12.

When solving for the second coordinate in the solution of a system of equations, use the simpler equation — the one with the smaller exponents — to avoid introducing extraneous solutions.

The answer is (–1, 9), (2, –3).

Solve for y in the second equation (you get y = 5 – 4x) and substitute the equivalent of y into the equation of the parabola: 5 – 4x = 4x 2 – 8x – 3. Move all the terms to the right and factor the equation: 0 = 4x 2 – 4x – 8 = 4(x 2 – x – 2) = 4(x + 1)(x – 2).

Using the multiplication property of zero, you find that x = –1 or x = 2. When x = –1 in the equation of the line, 4(–1) + y = 5; –4 + y = 5; y = 9. And substituting x = 2 in the equation of the line, 4(2) + y = 5; 8 + y = 5; y = –3.

How can you find the intersection, in this case 2 intersections of two functions like 9.2+0.9x and 7.0*1.10 x without using a graph? One is linear and one is exponential so no division or logs will give the answer, but you can graph it and find the intersections so surely there’s a way to do it algebraically. Anyone have an idea? Thanks.

One way to numerically approximate the solutions to a given equation is to use the Newton-Raphson Method. This method involves using basic calculus.

In this case, we’ll pick an arbitrary starting point. The method converges faster if you are able to use a graph to pick a value close to the actual answer, but you can start with any number you want. I’ll pick x_1 = 2. Now begins the iterative part of the process. First, we’ll establish the polynomial to represent the problem:

9.2 + 0.9x = 7 * 1.1 x

9.2/7 + (0.9/7)x = 1.1 x

9.2/7 + (0.9/7)x – 1.1 x = 0

f(x) = 9.2/7 + (0.9/7)x – 1.1 x

Next we take the derivative of this function:

f'(x) = 0.9/7 – ln(1.1) * 1.1 x

Now, using the x_1 = 2 established above, we have:

x_2 = x_1 – f(x_1)/f'(x_1) = 2 – [9.2/7 + (0.9/7)(2) – 1.1 2 ]/[0.9/7 – ln(1.1) * 1.1 2 ] ≈ -25.2856

x_3 = x_2 – f(x_2)/f'(x_2) = -25.2856 – [9.2/7 + (0.9/7)(-25.2856) – 1.1 -25.2856 ]/[0.9/7 – ln(1.1) * 1.1 -25.2856 ] ≈ -8.3993

x_4 = x_3 – f(x_3)/f'(x_3) ≈ -5.8959

x_5 = x_4 – f(x_4)/f'(x_4) ≈ -5.7092

And so on. As you can see, the method is converging to the actual answer. A graphical inspection reveals that the two functions do indeed intersect at around -5.71. A different initial value, such as x_1 = 7, may converge to the other intersection of the functions at around 10.05.

So you set it up as a single function equal to 0, find the derivative, then use some arbitary number to begin doing this process of X-f(x)/f'(x) and then use the answer as the new X in it to get a closer answer. Then repeating this until the new number you get is higher than the last as would be the next number of 5.764497(if I did it right) from what you started. This then tells you the point is between these two values. Do you then pick a number between those two and continue to get a new min and max values and repeat? In a sense this is picking points and testing it until you get one that’s right?

How to algebraically find the intersection of two lines

Steps for finding the intersection of the line and plane

If a line and a plane intersect one another, the intersection will be a single point, or a line (if the line lies in the plane).

How to algebraically find the intersection of two lines

I create online courses to help you rock your math class. Read more.

To find the point of intersection, we’ll

substitute the values of . x. . y. and . z. from the equation of the line into the equation of the plane and solve for the parameter . t.

take the value of . t. and plug it back into the equation of the line

This will give us the coordinates of the point of intersection.

The intersection of a line and a plane will either be a single point or a line

How to algebraically find the intersection of two lines

How to algebraically find the intersection of two lines

Take the course

Want to learn more about Calculus 3? I have a step-by-step course for that. 🙂

Intersection of a plane and a line given by parametric equations

Example

Find the point where the line intersects the plane.

The line is given by . x=-1+2t. . y=4-5t. and . z=1+t.

The plane is given by . 2x-3y+z=3.

Our first step is to plug the values for . x. . y. and . z. given by the equation of the line into the equation of the plane.

Now we’ll plug the value we found for . t. back into the equation of the line.

How to algebraically find the intersection of two lines

If a line and a plane intersect one another, the intersection will be a single point, or a line (if the line lies in the plane).

Putting these values together, we can say the point of intersection of the line and the plane is the coordinate point

If we want to double-check ourselves, we can plug this coordinate point back into the equation of the plane.

Since . 3=3. is true, we know that the point we found is a true intersection point with the plane.

Have you heard of point of intersection concept in mathematics? If no, don’t panic. Here, we will discuss the point of intersection in detail and how to calculate it either graphically or algebraically. Also, the formula is applicable to a variety of areas like businesses, finance, study, construction, or physics etc.

Have you ever noticed the traffic signal on a road? This is the example of point of intersection that will appear at the point when two roads are meeting up at a point. In mathematics, point of intersection is the point where two lines or curves generally meet.The value of two curves would be same significantly and it can be used at multiple places.

Take another example, if we wanted to represent the revenue of a Company against the costs then point of intersection would define the situation where revenue and costs are significantly the same. Most of the times, this is the breakeven point for a Company. The point can be calculated either graphically or algebraically.

Draw the graph of two equations and see where they will intersect visually. This is not a tough job but can be completed quickly with a deep understanding and practice. In most of the examples, you could analyze that graph is the best technique to find the point of intersection with accuracy.

Sometimes, there are the situation when this is not possible to find the point of intersection graphically then how can you solve the equation. The answer is you can do it algebraically. Solve the equations find the values of x coordinated that would point of intersection for both the equations.

Point Gradient Formula

For a line, the ratio of vertical change to the horizontal change is defined through a point i.e. named as the point of gradient or we can name it as the derivative as well. In brief, the gradient of a line will be rise divided by the run – rise/run. If m is the gradient point across a line then point gradient formula in mathematics could be given as –

Point Slope Form Formula

The other popular format for straight line equations is point slope formula. For this purpose, you need to find out the values (x1, y1) and a slope m. Further, plug the values into the formula –

Where,
m is the slope of the line.
x1 is the co-ordinates of x-axis.
y1 is the co-ordinates of y-axis.

Don’t scare of subscripts but they are just intended to indicate the points given to you. If you have the generic values for x and y coordinates then it can be directly plugged into the formula to calculate the final output. If you will calculate the values calculated from the slope-intercept form and the point slope form then they are exactly the same.

So, this is your choice which method are you planning to use and which technique suits you the most. Practice the technique and apply it as per your convenience for next mathematics problem.

You have to line segments and you want to know if they intersect. I’ll give you an algorithm how to do it.

Test cases

First of all, we should think about how lines can be arranged:

How to algebraically find the intersection of two lines

How to algebraically find the intersection of two lines

How to algebraically find the intersection of two lines

How to algebraically find the intersection of two lines

How to algebraically find the intersection of two lines

How to algebraically find the intersection of two lines

How to algebraically find the intersection of two lines

How to algebraically find the intersection of two lines

How to algebraically find the intersection of two lines

How to algebraically find the intersection of two lines

How to algebraically find the intersection of two lines

How to algebraically find the intersection of two lines

How to algebraically find the intersection of two lines

How to algebraically find the intersection of two lines

Bounding boxes

You can draw boxes around line segments such that the edges of the boxes are in parallel to the coordinate axes:

How to algebraically find the intersection of two lines

Two line segments with their bounding boxes

With this image in mind, it is obvious that the bounding boxes need to intersect if the lines should intersect. At this point you have to make a decision: If the endpoint of one line is on the other line, is this an intersection? I think so. If two lines have at least one point in common, they intersect. If two bounding boxes have at least one point in common, they intersect.

It is much easier to check if two bounding boxes intersect. It’s simply:

If you have difficulties to understand why this works, take a look at this great animation for this formula.

The algorithm

How to algebraically find the intersection of two lines

Flowchart how to check if two lines intersect

Looks quite simple, doesn’t it?

Cross product

Well, you might notice that you need to check if one line intersects with a given line segment. To check this, you have to understand one cool idea:

You can definie a cross product for points:

This cross product has one nice characteristics:

\(a \times_P b = 0 \Leftrightarrow a\) and \(b\) are on one line through origin

You can verify this. If you take two points on a line through origin, they have the same slope \(\frac<\Delta y><\Delta x>\) :

Ok, now you can check if a point is on a line:

The second cool characteristic of the cross product is that it can be used to determine if a point b is left or right of the line through the origin and a point a:

When we have one line \(a\) through the origin and one line segment \(b\) , you can check if \(b\) crosses \(a\) by checking if the end points of \(b\) are on different sides of \(a\) :

Now you have everything you need:

By the way, testcase F5 is the only reason why you need doBoundingBoxesIntersect(box1, box2) . All other tests still pass if you remove this function.

Where do two line segments intersect?

When you know that two line segments intersect, you can also calculate the intersection. The intersection could be a line or only a point.

I did this with JavaScript:

Ihr Browser kann leider keine eingebetteten Frames anzeigen. Die Seite ist hier.

This is the code that checks for line segments:

The complete, tested code is on GitHub. Here is the most important part:

Addendum

Some notes for me:

  • Writing Tests first is worth the effort. I guess it finally saved me some time and it gives me some confidence that my code works.
  • I should update my system. I’m still using Ubuntu 10.04.4 LTS. Especially, I have Eclipse 3.5.2. This means I could not try EclEmma to test my code coverage ☹
  • LaTeX is great. I’ve created all images with LaTeX and it was quite fast after I got the first one. Here is the LaTeX source.

edit: I now have a more modern system. So I was able to use EclEmma, which works fine. And I have 100% branch code coverage for this part of code ☺

How to algebraically find the intersection of two lines

To determine whether two lines are parallel, intersecting, skew or perpendicular, we will need to perform a number of tests on the two lines.

How to algebraically find the intersection of two lines

I create online courses to help you rock your math class. Read more.

Given two lines,

then the lines are

parallel if the ratio equality is true.

intersecting if the lines are not parallel or if you can solve them as a system of simultaneous equations.

perpendicular if the lines are intersecting and their dot product is . 0.

skew if the lines are not parallel and not intersecting.

How to determine if two lines are parallel, intersecting (or perpendicular specifically), or skew

How to algebraically find the intersection of two lines

How to algebraically find the intersection of two lines

Take the course

Want to learn more about Calculus 3? I have a step-by-step course for that. 🙂

Test for perpendicular first, then intersecting (then perpendicular if intersecting), and then skew

Example

Say whether the lines are parallel, intersecting, perpendicular or skew.

We’ll start by testing the lines to see if they’re parallel by pulling out the coefficients

Since . 5/3\neq1/2\neq-1/2. we know the lines are not parallel.

Because they’re not parallel, we’ll test to see whether or not they’re intersecting. We’ll set the equations for . x. . y. and . z. from each line equal to each other. If we can find a solution set for the parameter values . s. and . t. and this solution set satisfies all three equations, then we’ve proven that the lines are intersecting.

Setting . x_1=x_2. we get

[1] . t=\frac15+\frac35s.

Setting . y_1=y_2. we get

[2] . -3+2t=3+4s.

Plugging [1] into [2] gives

[3] . s=-2.

Plugging [3] into [1] gives

[4] . t=-1.

Setting . z_1=z_2. we get

[5] . 1+t=3-2s.

Plugging [3] and [4] into [5] gives

Since . 0\neq7. the lines are not intersecting.

Because . L_1. and . L_2. are not parallel and not intersecting, by definition they must be skew.

How to algebraically find the intersection of two lines

Two lines are intersecting if the lines are not parallel or if you can solve them as a system of simultaneous equations.

In the previous example, we didn’t test for perpendicularity because only intersecting lines can be perpendicular, and we found that the lines were not intersecting. If we had found that . L_1. and . L_2. were in fact perpendicular, we would have needed to test for perpendicularity by taking the dot product, like this:

Since the dot product isn’t . 0. we’ve proven that the lines are not perpendicular.

Given points A and B corresponding to line AB and points P and Q corresponding to line PQ, find the point of intersection of these lines. The points are given in 2D Plane with their X and Y Coordinates.

Recommended: Please try your approach on first, before moving on to the solution.

First of all, let us assume that we have two points (x1, y1) and (x2, y2). Now, we find the equation of line formed by these points.

  1. a1x + b1y = c1
  2. a2x + b2y = c2

We have to now solve these 2 equations to find the point of intersection. To solve, we multiply 1. by b2 and 2 by b1
This gives us,
a1b2x + b1b2y = c1b2
a2b1x + b2b1y = c2b1

This gives us the value of x. Similarly, we can find the value of y. (x, y) gives us the point of intersection.

  • min (x1, x2) <= x <= max (x1, x2)
  • min (y1, y2) <= y <= max (y1, y2)

The pseudo code for the above implementation:

These can be derived by first getting the slope directly and then finding the intercept of the line.

Two or more lines intersect when they share a common point.

If two lines share more than one common point, they must be the same line. If two lines in the same plane share no common point, they must be parallel.

Same line Parallel lines
Line m and n share points A and B so they are the same line. In the same plane, lines m and n share no common points, so they are parallel.

Coordinate geometry and intersecting lines

In coordinate geometry, the graphs of lines can be written as equations. Points of intersection can be found using the equations of the lines.

Find the point of intersection for the lines whose equations are,

y = 3x – 2
y = -x + 6

One method to find the point of intersection is to substitute the value for y of the 2 nd equation into the 1 st equation and solve for the x-coordinate.

-x + 6 = 3x – 2
-4x = -8
x = 2

Next plug the x-value into either equation to find the y-coordinate for the point of intersection.

y = 3×2 – 2 = 6 – 2 = 4

So, the lines intersect at (2, 4).

How to algebraically find the intersection of two lines

Intersecting lines and angles

Angles are formed when two or more lines intersect.

How to algebraically find the intersection of two lines

In the figure above, MP and NQ intersect at point O forming four angles that have their vertices at O.

    are congruent so, ∠MOQ≅∠NOP and ∠MON≅∠QOP.
  • Angles ∠MOQ and ∠QOP, and angles ∠NOP and ∠QOP form a linear pair, so ∠MOQ + ∠QOP = 180° and ∠NOP + ∠QOP = 180°.

Forming a plane

The intersection of two lines forms a plane.

In the figure above, line m and n intersect at point O. Together, lines m and n form plane p.

If we consider two or more equations together we have a system of equations. A solution to a system of equations in \(x\) and \(y\) is a pair of values \(a\) and \(b\) for \(x\) and \(y\) that make all of the equations true.

Do you think such a solution exists for the system of equations in part (b)? Explain.

Commentary

The purpose of this task is to introduce students to systems of equations. It takes skills and concepts that students know up to this point, such as writing the equation of a given line, and uses it to introduce the idea that the solution to a system of equations is the point where the graphs of the equations intersect (assuming they do). This task does not delve deeply into how to find the solution to a system of equations because it focuses more on the student’s comparison between the graph and the system of equations.

The language in the task stem states that a solution to a system of equations is a pair of values that make all of the equations true. The solution shortens this to “satisfying” the equations–this is a more succinct way of saying it, but students may not know that “the ordered pair of values \((a,b)\) satisfies an equation” means “\(a\) and \(b\) make the equation true when \(a\) is substituted for \(x\) and \(b\) is substituted for \(y\) in the equation.” Many people, books, and assessments talk about pairs of values “satisfying” an equation, so it would be helpful to students to have the meaning of this word made explicit.

Solutions

    Constructing a set of axes, we can first locate the two given points, \((1,4)\) and \((0,-1)\), to create our first line. Our second line can be any other line that passes through \((1,4)\) but not \((0,-1)\), so there are many possible answers. Below is one possible construction:

How to algebraically find the intersection of two lines

Graphically, we can start at the point \((0,-1)\) and then count how many units we go up divided by how many units we then go right to get to the point \((1,4)\), as in the diagram below.

How to algebraically find the intersection of two lines

This gives a slope of \(m=\displaystyle\frac<5><1>=5\).

Algebraically, we can find the difference between the \(y\)-coordinates of the two points, and divide it by the difference between the \(x\)-coordinates. $$m=\frac<4-(-1)><1-0>=5.$$ Because the \(y\)-intercept of this line is -1, we have \(b=-1\). So, the equation of our first line is \(y=5x-1\).

We can reason in a similar way for our second line. Graphically, we see our second line contains the point \((0,6)\), so we can start at the point \((0,6)\) and then count how many units we go down divided by how many units we then go right to get to the point \((1,4)\), as in the diagram below.

How to algebraically find the intersection of two lines

This gives a slope of \(\displaystyle m=\frac<-2><1>=-2\).

We can also find the slope algebraically: $$m=\frac<4-6><1-0>=-2.$$ Because we have a \(y\)-intercept of 6, \(b=6\). So, the equation of our first line is \(y=-2x+6\).

Using this idea that a solution to a system of equations is a pair of values that makes both equations true, we decide that our system of equations does have a solution, because

  • The coordinates of every point on a line satisfy its equation, and
  • The point \((1,4)\) lies on both lines.

We can confirm that \((1,4)\) is our system’s solution by substituting \(x=1\) and \(y=4\) into both equations: $$4=5(1)-1$$ and $$4=-2(1)+6.$$

Most of us must find intersection of two linear straight lines with pen and paper during school days. In this article, we will see how to solve it with Excel.

  • First we need the equations of the two lines.
  • Then, since at the point of intersection, the two equations will have the same values of x and y, we set the two equations equal to each other. This gives an equation that we can solve for x
  • We substitute that x value in one of the line equations and solve it for y

Where
m1: Gradient or Slope of line1,
c1: Intercept of line1,
m2: Gradient or Slope of line2,
c2: Intercept of line2

Let the point of intersection be s and t. They must satisfy the above 2 equations.

t=m1s+c1 and t=m2s+c2 . Then m1s+c1=m2s+c2
s= (c2-c1)/(m1-m2) , t=m1*s+c1

For example : I take the following equations :

y1 = 2.5×1 – 0.5
y2=-0.5×1+1.67

Based on the above equations i created data for y1 and y2 using same values on X1.

To find the intersection point :

Apply the following formulae :

and Y =SLOPE(y-values1,x-values1)*X+INTERCEPT(y-values1,x-values1)

Excel : Intersection of 2 linear Straight Lines

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Suppose f ( x ) and g ( x ) are two functions that take a real number input, and output a real number.

Then the intersection points of f ( x ) and g ( x ) are those numbers x for which f ( x ) = g ( x ) .

Sometimes the exact values can be easily found by solving the equation f ( x ) = g ( x ) algebraically.

Example 1:

What are the points of intersection of the functions f ( x ) and g ( x ) if f ( x ) = x + 6 and g ( x ) = − x ?

The intersection points of f ( x ) and g ( x ) are those numbers x for which f ( x ) = g ( x ) .

That is, x + 6 = − x .

x + 6 = − x 2 x + 6 = 0 2 x = − 6 x = − 3

Now, you can use the value of x to find the corresponding y -coordinate of the point of intersection.

Substitute the value of x in any of the two functions.

The equations can also be solved graphically by plotting the two functions on a coordinate plane and identifying the point of intersection of the two.

In other cases, the exact values can be hard to find. You may need to use technology to estimate them.

Example 2:

Find the point(s) of intersection of the two functions.

f ( x ) = |   x − 5   | g ( x ) = log x

Here, solving algebraically is not so easy .The solutions to the equation | x − 5 | = log x are not nice-looking rational numbers.

Graph the functions on a coordinate plane .

You can use a graphing utility to find that the coordinates of the intersection points are approximately ( 4.36 , 0.64 ) and ( 5.76 , 0.76 ) .

How to algebraically find the intersection of two lines

One of common tasks in algebra homework is to solve a system of two linear equations. Along with other methods, we can do it graphically. This section is devoted to intersection of two straight lines on a plane. In other words, let’s talk about solving a system of two linear equations. As we discussed earlier, a straight line on a plane is described by the following equation:

This equation sets a certain dependence between coordinates $x$ and $y$ of the point lying on this line. So, if we want to discuss intersection of two lines, we’ll have to deal with two equations of such kind.

Consider the following system:

Each of the two linear equations represents a straight line, so let’s graph them.

The first equation has y -intercept b=3 , thus we have point (0,3) .
Slope m=2=\frac<2> <1>, thus the second point is (1,5) . How did we get this? Recall the rise-over-run formula for the slope:

Rise means increase in y coordinate, run goes for increase in x coordinate. Thus, in our case when x “runs” for 1 , y “rises” for 2 . So, (0,3) \rightarrow (0+1,3+2)=(1,5) . That’s our second point.

Let’s take a look at the second equation. We should solve it for y first:

x-2y = 0 \rightarrow -2y=-x \rightarrow y=0.5x

Ok, now we finally have this equation in slope-intercept form, y -intercept b=0 (that’s because the corresponding term is absent in the equation, we have just 0.5x in the right part), slope m=0.5=\frac<1> <2>. Thus, the points are (0,2) , (1,0.5) .

Check also video tutorial on our youtube channel:

How to algebraically find the intersection of two lines

As we have a pair of points for each line, we can easily graph them:

How to algebraically find the intersection of two lines

We’ve already discussed that a solution of single linear equation is every point lying on the line which represents that equation. If we consider a system of two equations, the solution of such system will be the point lying on both lines at the same time. We can see from the graph that our lines have only one common point, they intersect at (-2,-1) . Therefore, the solution of our systems of equations is x=-2, y=-1 .

In general, not every system of two linear equations has single solution like in our example. Our two lines intersected, but that’s not always the case. Two lines on a plane can either be parallel or coincide, or intersect at a single point.
Let’s take a look at some examples.

Two parallel lines

Consider the following pair of lines:

The first one is the same as in previous example. Consider the second equation. The y -intercept is b=1 , thus we have the point (0,1) . The slope m is the same as for the first line: m=2 , therefore the second point is (1,3) . Two points are enough to graph a line, as we know. Look at the graph:

How to algebraically find the intersection of two lines

Here are two parallel lines which never intercept. Hence, the system of equations has no solution at all.
As we know, slope m shows how steep the line is. If two lines have the same slopes they are of the same steepness and thus parallel, like in our case.

Coinciding lines

At first let’s solve the second equation for y :

4x-2y+6 = 0 \rightarrow 4x+6=2y

Diving by 2 both parts of equation we obtain:

or in usual form:

Now our system of equations looks like this:

Thus, the two equations of the system are the same. It means that when drawn on the plane they will coincide. Every point of the two coinciding lines is a solution of the system. In this case it’s said that a system has infinite number of solutions.

Summing up. If you’re asked to solve a system of two linear equations, this means you are dealing with two lines. Solution of a system is a set of points at which these two lines intersect. Three cases are possible:

– two lines intersect at just one point. The system has unique solution.

– the lines are parallel, they never meet and have no common points, thus the system has no solutions.

– the lines coincide, i.e. they are represented by identical equations. Each point of one line also belongs to another. In this case the system has infinitely many solutions.

That’s what you need to know about intersection of two lines on a plane and their connection with system of two linear equations. Still have questions? Concerned with any other math, physics, chemistry problems in your homework? Just let us know, we’re ready to help.