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 clear, from the given table, that the line passes through the points, $\left(-1,-5\right)$, $\left(1,1\right)$ and $\left(3,7\right)$.

Then, let $\left(h,k\right)=\left(-1,-5\right)$, $\left(a,b\right)=\left(1,1\right)$ and $\left(c,d\right)=\left(3,7\right)$.

Put $\left(a,b\right)=\left(1,1\right)$and $\left(c,d\right)=\left(3,7\right)$in $m=\frac{d-b}{c-a}$to get,

$\begin{array}{l}m=\frac{7−1}{3−1}\\ =\frac{6}{2}\\ =3\end{array}$

So, $m=3$

So, the slope of the required line is $3$.

Put $m=3$and $\left(h,k\right)=\left(-1,-5\right)$in $y-k=m\left(x-h\right)$to get,

$y-\left(-5\right)=3\left(x-\left(-1\right)\right)$

$y+5=3\left(x+1\right)$ (Simplify)

$y+5=3x+3$ (Distributive property)

$y+5-5=3x+3-5$ (Subtract 5 from both sides)

$y=3x-2$ (Simplify)

So, the required equation of the straight line in slope-intercept form is $y=3x-2$