Question

Find the equation of line passing through the point $\left( -6,10 \right)$ and perpendicular to the straight line $7x+8y=5$.

Answer 1

Hint: Use the fact that the product of slopes of two perpendicular lines is $-1$ to find the equation of line. Use one point form of line to write the equation of line with a given slope passing through a point.

Complete step-by-step answer:
We have to find the equation of line passing through the point $\left( -6,10 \right)$ and perpendicular to the straight line $7x+8y=5$.
We will firstly evaluate the slope of the given line.
We know that the slope of the line of the form $ax+by=c$ is $\dfrac{-a}{b}$.
Substituting $a=7,b=8,c=5$ in the above formula, the slope of line $7x+8y=5$ is $\dfrac{-7}{8}$.
Let’s assume that the slope of line passing through the point $\left( -6,10 \right)$ and perpendicular to the straight line $7x+8y=5$ is $m$.
We know that the product of slopes of two perpendicular lines is $-1$.
Thus, we have $m\left( \dfrac{-7}{8} \right)=-1$.
$\Rightarrow m=\dfrac{8}{7}$
We know that the equation of line passing through point $\left( a,b \right)$ having slope $m$ is $y-b=m\left( x-a \right)$.
Substituting $m=\dfrac{8}{7},a=-6,b=10$ in the above formula, we get the equation of line as $y-10=\dfrac{8}{7}\left( x+6 \right)$.
Simplifying the above equation, we get $7y-70=8x+48$.
$\Rightarrow 7y=8x+118$

Hence, the equation of line passing through the point $\left( -6,10 \right)$ and perpendicular to the straight line $7x+8y=5$ is $7y=8x+118$.

Note: It’s necessary to use the fact that the product of slopes of two perpendicular lines is $-1$ and then use point slope form to write the equation of line. We won’t be able to write the equation of line without using these facts. Any equation of line represents the relation between points lying on the line. Point slope form of line gives a relation between the slope of line and any general point lying on the line.