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 slopes of parallel lines are equal.

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

The given line is parallel to $2x+5y=10$ and goes through the point $\left(4,-3\right)$.

The required line is parallel to the line $2x+5y=10$.

Converting this straight-line equation to slope-intercept form,

$2x+5y=10$ (Given equation)

$2x+5y-2x=10-2x$ (Subtract $2x$ from both sides)

$5y=-2x+10$ (Simplify and rearrange)

$\frac{5y}{5}=\frac{-2x+10}{5}$ (Divide both sides by 5)

$y=-\frac{2}{5}x+2$(Simplify)

Comparing with $y=mx+c$, $m=-\frac{2}{5}$ and $c=2$.

So, slope of the line $2x+5y=10$ and thus the required line is $m=-\frac{2}{5}$.

Put $m=-\frac{2}{5}$ and $\left(h,k\right)=\left(4,-3\right)$ in $y-k=m\left(x-h\right)$ to get,

$y-\left(-3\right)=-\frac{2}{5}\left(x-4\right)$

$y+3=-\frac{2}{5}\left(x-4\right)$ (Simplify)

$5\left(y+3\right)=-\frac{2}{5}\left(x-4\right)\left(5\right)$ (Multiply both sides by 5)

$5y+15=-2\left(x-4\right)$ (Simplify)

$5y+15=-2x+8$ (Simplify)

$2x+5y+15=-2x+8+2x$ (Add $2x$ to both sides)

$2x+5y+15=8$ (Simplify)

$2x+5y+15-15=8-15$ (Subtract 15 from both sides)

$2x+5y=-7$ (Simplify)

So, $2x+5y=-7$ is the equation of the required line.

Graphing the obtained equation of the straight line,