Q40. - Algebra 2

Question

Write an equation in slope-intercept form of the line that passes through $\left(-\mathbf{2},\mathbf{10}\right)$, $\left(\mathbf{2},\mathbf{2}\right)$ and $\left(\mathbf{4},-\mathbf{2}\right)$.

Step-by-Step Solution

Answer

The equation of the required straight line in slope-intercept form is $y=-2x+6$.

Step 1 – State the concept

The slope intercept form of a straight-line equation is $y=mx+c$ where $m$ is the slope and $c$ is the y-intercept.

 The slope of a line passing through $\left(a,b\right)$ and $\left(c,d\right)$ is $m=\frac{d-b}{c-a}$.

 The equation of a straight-line having slope $m$ and passing through the point $\left(h,k\right)$ is given as $y-k=m\left(x-h\right)$

Step 2 – List the given data

It is given that the line passes through the points, $\left(-2,10\right)$, $\left(2,2\right)$ and $\left(4,-2\right)$.

Then, let $\left(h,k\right)=\left(-2,10\right)$, $\left(a,b\right)=\left(2,2\right)$ and $\left(c,d\right)=\left(4,-2\right)$.

Step 3 – Determine the slope

Put $\left(a,b\right)=\left(2,2\right)$ and $\left(c,d\right)=\left(4,-2\right)$ in $m=\frac{d-b}{c-a}$ to get,

$\begin{array}{l}m=\frac{-2-2}{4-2}\\ =\frac{-4}{2}\\ =-2\end{array}$

So, $m=-2$

 So, the slope of the required line is $-2$.

Step 4 – Write the equation

Put $m=-2$ and $\left(h,k\right)=\left(-2,10\right)$ in $y-k=m\left(x-h\right)$ to get,

 $y-10=-2\left(x-\left(-2\right)\right)$

 $y-10=-2\left(x+2\right)$  (Simplify)

$y-10=-2x-4$   (Distributive property)

$y-10+10=-2x-4+10$   (Add 10 to both sides)

 $y=-2x+6$   (Simplify)

 So, the required equation of the straight line in slope-intercept form is $y=-2x+6$.