Equations of straight lines

In a nutshell

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

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

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

Procedure

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Example 1

​A line has a point $$(3,4)$$​ and a gradient of $$2$$​. Find the equation of the line in the form of $$y=mx+c$$.

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​Label the gradient and the coordinates.​

​​$$m=$$​$$2$$​​, $$x_1=$$​$$3$$​​, $$y_1=4$$

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Substitute $$y_1$$, $$m$$ and $$x_1$$ in the equation $$y-y_1=m(x-x_1)$$.

​​$$y-4=2(x-3)$$

Rearrange to give $$y=mx+c$$.

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

Finding $$y=mx+c$$ using two points​

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

Procedure

Example 2

​A line intersects two points, $$P_1$$​​ with coordinates  $$(-2, 5)$$​ and $$P_2$$​ with coordinates $$(-6,1)$$​. Find the equation of the line in the format $$y=mx+c$$​.

Label both points.

​$$x_1=-2, y_1=5, x_2=-6,y_2=1$$

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Find the gradient $$m$$.

$$\begin{aligned}m&=\dfrac{y_2-y_1}{x_2-x_1}\\ \\m&=\dfrac{1-5}{-6-(-2)}\\ \\m&=\dfrac{-4}{-4}\\ \\m&=1\\\end{aligned}$$

Solve for $$y=mx+c$$.

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

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 $$x$$ or $$y$$​ in one equation with an answer from the other. This works mathematically since the point where those lines intersect means that the $$x$$ and $$y$$ values in both equations must be equal.

Example 3

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

Substitute $$y$$​ in one equation.

$$\begin{aligned}5x-2y+12&= 0\\2x+5&=y\\5x-2(2x+5)+12&=0\\\end{aligned}$$

Solve for $$x$$​

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

Use this $$x$$ value to solve for the corresponding $$y$$​ value to find point $$A$$​.​​​​​

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

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