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Equations of straight lines

Equations of straight lines

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Explainer Video

Tutor: Daniel

Summary

Equations of straight lines

In a nutshell

You can find out the equation of a straight line, y=mx+cy=mx+cy=mx+c, using gradients and points. The equation of a straight line can be found by using a pair of coordinates (x1,y1)(x_1,y_1)(x1​,y1​)​ and the gradient mmm. You can use an equation, y−y1=m(x−x1)y-y_1=m(x-x_1)y−y1​=m(x−x1​), to find the exact equation for y=mx+cy=mx+cy=mx+c​. You can also use equations to find where lines intersect.



Finding y=mx+cy=mx+cy=mx+c using the gradient and a point​

If you know the gradient of a line, mmm​, and you know a point on that line, (x1,y1)(x_1,y_1)(x1​,y1​), then you can work out the equation of that line. All you have to do is use the formula y−y1=m(x−x1) y-y_1=m(x-x_1)y−y1​=m(x−x1​). This will allow you find the line's equation.


Procedure

1.

Label the gradient and the coordinates of the point on the line.

2.

Substitute into the equation y−y1=m(x−x1)y-y_1=m(x-x_1)y−y1​=m(x−x1​).​

3.

Solve for y=mx+cy=mx+cy=mx+c​.

​

Example 1

​A line has a point (3,4)(3,4)(3,4)​ and a gradient of 222​. Find the equation of the line in the form of y=mx+cy=mx+cy=mx+c.

​

​Label the gradient and the coordinates.​

​​m=m=m=​222​​, x1=x_1=x1​=​333​​, y1=4y_1=4y1​=4

​

Substitute y1y_1y1​, mmm and x1x_1x1​ in the equation y−y1=m(x−x1)y-y_1=m(x-x_1)y−y1​=m(x−x1​).

​​y−4=2(x−3)y-4=2(x-3)y−4=2(x−3)


Rearrange to give y=mx+cy=mx+cy=mx+c.

​​​​y−4=2x−6y=2x−2‾\begin{aligned}y-4 &= 2x-6\\&\underline{y =2x-2}\\\end{aligned}y−4​=2x−6y=2x−2​​



Finding y=mx+cy=mx+c y=mx+c using two points​

To find the equation of a line using two points (x1,y1)(x_1, y_1)(x1​,y1​)​ and (x2,y2)(x_2,y_2)(x2​,y2​), first you find the gradient mmm​ using the equation from the previous lesson, m=y2−y1x2−x1m=\dfrac{y_2-y_1}{x_2-x_1}m=x2​−x1​y2​−y1​​. Once you find out mmm, you can then take the (x1,y1)(x_1, y_1)(x1​,y1​) coordinates and solve as above.


Procedure

1.

Label the coordinates (x1,y1)(x_1,y_1)(x1​,y1​) and (x2,y2)(x_2,y_2)(x2​,y2​).

2.

Find out the gradient mmm.

3.

Rearrange for y=mx+cy=mx+cy=mx+c.


Example 2

​A line intersects two points, P1P_1P1​​​ with coordinates  (−2,5)(-2, 5)(−2,5)​ and P2P_2P2​​ with coordinates (−6,1)(-6,1)(−6,1)​. Find the equation of the line in the format y=mx+cy=mx+cy=mx+c​.


Label both points.

​x1=−2,y1=5,x2=−6,y2=1x_1=-2, y_1=5, x_2=-6,y_2=1x1​=−2,y1​=5,x2​=−6,y2​=1

​​

Find the gradient mmm.

m=y2−y1x2−x1m=1−5−6−(−2)m=−4−4m=1\begin{aligned}m&=\dfrac{y_2-y_1}{x_2-x_1}\\ \\m&=\dfrac{1-5}{-6-(-2)}\\ \\m&=\dfrac{-4}{-4}\\ \\m&=1\\\end{aligned}mmmm​=x2​−x1​y2​−y1​​=−6−(−2)1−5​=−4−4​=1​


Solve for y=mx+cy=mx+cy=mx+c.

​y−y1=m(x−x1)y−5=1(x−(−2))y=x+2+5y=x+7‾\begin{aligned}y-y_1&=m(x-x_1)\\y-5&=1(x-(-2))\\y&= x +2+5\\&\underline{y=x+7}\\\end{aligned}y−y1​y−5y​=m(x−x1​)=1(x−(−2))=x+2+5y=x+7​​,

 


Find the intersection between two lines

Finally, sometimes you will be given two line equations and asked to find a point where those lines intersect. This can be solved like simultaneous equations, where you substitute xx x or yyy​ in one equation with an answer from the other. This works mathematically since the point where those lines intersect means that the xxx and yyy values in both equations must be equal.


Example 3

The lines y=2x+5y=2x+5y=2x+5​ and 5x−2y+12=05x-2y+12=05x−2y+12=0​ intersect at the point AAA​. Find the coordinates of AAA​.


Substitute yyy​ in one equation.

5x−2y+12=02x+5=y5x−2(2x+5)+12=0\begin{aligned}5x-2y+12&= 0\\2x+5&=y\\5x-2(2x+5)+12&=0\\\end{aligned}5x−2y+122x+55x−2(2x+5)+12​=0=y=0​


Solve for xxx​

​5x−4x−10+12=0x+2=0x=−2\begin{aligned}5x-4x-10+12&=0\\x+2&=0\\x&=-2\\\end{aligned}5x−4x−10+12x+2x​=0=0=−2​


Use this xx x value to solve for the corresponding yyy​ value to find point AAA​.​​​​​

y=2x+5y=2(−2)+5y=−4+5y=1∴ A=(−2,1)‾\begin{aligned}y&=2x+5\\y&=2(-2)+5\\y&=-4+5\\y&=1\\&\underline{\therefore\space A =(-2,1)}\\\end{aligned}yyyy​=2x+5=2(−2)+5=−4+5=1∴ A=(−2,1)​​​​​​

​​ 

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Equation of a straight line: y = mx + c

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Equation of a straight line: y = mx + c

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FAQs - Frequently Asked Questions

How do you find the equation of a line?

Using a point (x1,y1) and the gradient m, you can use the formula y-y1=m(x-x1) to find the equation of a line.

What are the three equations of a line?

y=mx+c, where m is the gradient of the line and c is the y-intercept. y-y1=m(x-x1) is the equation to find y=mx+c using one point (x1,y1) and the gradient m. Ax+By=C is the standard form of the equation, where A, B, and C are all integers and A is positive.

What are the 2 equations of a straight line?

The equation for a straight line is y=mx+c, where m is the gradient of the line and c is the y-intercept. y=c is where the line hits the y-axis, or where x=0.

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