91Ó°ÊÓ

The y-intercept is the value of y when $x=0$.

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

The slope of a line perpendicular to a line having slope m is $-\frac{1}{m}$.

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 required line is perpendicular to $2x+5y=10$ and intersects its graph at its y-intercept.

The required line is perpendicular 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 $2x+5y=10$ is $m=-\frac{2}{5}$. Then, slope of the required line is $\frac{5}{2}$.

Put $x=0$ in $2x+5y=10$ to get,

$2\left(0\right)+5y=10$

$5y=10$ (Simplify)

$\frac{5y}{5}=\frac{10}{5}$ (Divide both sides by 5)

$y=2$ (Simplify)

So, the y-intercept is 2 and the point through which the required line passes is $\left(0,2\right)$.

Put $m=\frac{5}{2}$ and $\left(h,k\right)=\left(0,2\right)$ in $y-k=m\left(x-h\right)$ to get,

$y-2=\frac{5}{2}\left(x-0\right)$

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

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

$2y-4=5x$ (Simplify)

$2y-4-2y=5x-2y$ (Subtract $2y$ from both sides)

$5x-2y=-4$(Simplify and rearrange)

So, $5x-2y=-4$ is the equation of the required line.

Graphing the obtained equation of the straight line,