Question

Find the equation of the line passing through the point $\left( 2,3 \right)$ and perpendicular to the straight line $4x-3y=10$ .

Answer 1

Hint: To find the equation of the line passing through the point $\left( 2,3 \right)$ and perpendicular to the straight line $4x-3y=10$ , we have to consider this line in slope-intercept form as $y=mx+c$ . We have to change the equation of the given straight line into this slope-intercept form. We know that when two lines are perpendicular, the product of their slopes will be equal to -1. From this, we can get the slope of the required line. Now, to find c substitute the given point and the slope obtained in the slope-intercept form.

Complete step by step solution:
We have to find the equation of the line passing through the point $\left( 2,3 \right)$ and perpendicular to the straight line $4x-3y=10$ . Let us denote this line in slope-intercept form.
$\Rightarrow y=mx+c..\left( i \right)$
where m is the slope of the line and c is the y-intercept.
We are given the equation of the straight line.
 $4x-3y=10$
Let us change this equation in the slope-intercept form. We have to take 4x to the RHS.
$\Rightarrow -3y=-4x+10$
Let us take –y to the RHS.
$\begin{align}
  & \Rightarrow y=\dfrac{-4x+10}{-3} \\
 & \Rightarrow y=\dfrac{4}{3}x-\dfrac{10}{3} \\
\end{align}$
We can see the slope of this line, say ${{m}_{1}}=\dfrac{4}{3}$ . We are given that the required line is perpendicular to the above line. Hence, the product of their slopes will be equal to -1.
$\Rightarrow m{{m}_{1}}=-1$
Let us find the slope of the required line from the above equation.
$\begin{align}
  & \Rightarrow m=\dfrac{-1}{{{m}_{1}}} \\
 & \Rightarrow m=\dfrac{-1}{\dfrac{4}{3}}=-\dfrac{3}{4} \\
\end{align}$
Let us substitute the above value in the equation (i).
$\Rightarrow y=\dfrac{-3}{4}x+c...\left( ii \right)$
Now, we have to find the y-intercept, that is, c. We are given that the line passes through the point $\left( 2,3 \right)$ . Let us substitute this value of x and y in the equation (ii).
$\Rightarrow 3=\dfrac{-3}{4}\times 2+c$
Let us simplify the above equation and find c.
$\begin{align}
  & \Rightarrow 3=\dfrac{-3}{2}+c \\
 & \Rightarrow c=3+\dfrac{3}{2}=\dfrac{6}{2}+\dfrac{3}{2} \\
 & \Rightarrow c=\dfrac{9}{2} \\
\end{align}$
Now, we have to substitute this value of c in equation (ii).
$\Rightarrow y=\dfrac{-3}{4}x+\dfrac{9}{2}$
We can rearrange this by taking LCM in the RHS.
$\begin{align}
  & \Rightarrow y=\dfrac{-3}{4}x+\dfrac{18}{4} \\
 & \Rightarrow y=\dfrac{-3x+18}{4} \\
 & \Rightarrow 4y=-3x+18 \\
\end{align}$
Let us take -3x to the LHS.
$\Rightarrow 3x+4y=18$

Hence, the equation of the required line is $3x+4y=18$

Note: When two lines are perpendicular, the product of their slopes will be equal to -1. Similarly, when we are given with two parallel lines, their slopes will be equal. Students must never forget to substitute back the value of c in the equation of the required line.