91Ó°ÊÓ

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 $y-k=m\left(x-h\right)$

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 $\left(c,d\right)=\left(4,-2\right)$.

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 $-2$.

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 $y=-2x+6$.