Question

Find the equation of a line drawn perpendicular to the line $\frac{\mathrm{x}}{4}+\frac{\mathrm{y}}{6}=1$ through the point, where it meets the $Y-axis.$

Answer 1

Step1- Finding the slope of line:

Given equation is $\frac{\mathrm{x}}{4}+\frac{\mathrm{y}}{6}=1$

Multiplying $12$ on both sides,

We get,

$\text{⇒ 3x + 2y = 12}$

$$\begin{gathered} \text{⇒ 2y =-3x+ 12} \\ \Rightarrow y=\frac{-3}{2}x+6.........\left(1\right) \end{gathered}$$

From above equation, the $y$ -intercept is $6$ so the point on $x-axis$ will be $(0,6)$

From equation (1) the slope $m$ will be $\frac{-3}{2}$

Step2- Finding the equation of line :

Given that the required line is perpendicular to the given line $y=\frac{-3}{2}x+6$,

Then, the multiplication of their slopes will$-1$

$$\begin{gathered} \Rightarrow m\times \frac{-3}{2}=-1 \\ \Rightarrow m=\frac{2}{3} \end{gathered}$$

Equation of a Line is given by

$$\begin{gathered} \text{⇒}\left(\mathrm{y}-{\mathrm{y}}_1\right)=\mathrm{m}(\mathrm{x}-{\mathrm{x}}_1) \\ \Rightarrow (\mathrm{y}-6)=\frac{2}{3}(\mathrm{x}-0) \\ \Rightarrow 3(\mathrm{y}-6)=2\left(\mathrm{x}\right) \\ \Rightarrow 3\mathrm{y}-18=2\mathrm{x} \\ \Rightarrow 2\mathrm{x}-3\mathrm{y}+18=0 \end{gathered}$$

Therefore the required equation of line is $2x-3y+18=0$