Question

Find the equation of a line passing through the point $\left(2,4\right)$ and intersecting the line $2x+3y+6=0$ on the $X-axis$.

Answer 1

Step 1:Finding the another coordinate through which the line is passing:

Given equation of line is $2x+3y+6=0$ and it is passing through $X-axis$

So we can substitute the value $y=0$,

$\begin{array}{rcl}2x+3y+6& =& 0\\ & \Rightarrow & 2x+3\times 0+6=0\\ & \Rightarrow & 2x+6=0\\ & \Rightarrow & x=-\frac{6}{2}\\ & \Rightarrow & x=-3\end{array}$

Therefore, the required line is passing through $\left(2,4\right)$ and $\left(-3,0\right)$.

Step 2:Finding the equation of a line:

Equation of line passing through points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given as:

$y-y_1=\frac{y_2-y_2}{x_2-x_1}\left(x-x_1\right)$

Here, $\left(x_1,y_1\right)=(2,4)$ and $\left(x_2,y_2\right)=(-3,0)$

Equation of line is given as:

$\begin{array}{rcl}y-4& =& \frac{0-4}{-3-2}\left(x-2\right)\\ & \Rightarrow & y-4=\frac{-4}{-5}\left(x-2\right)\\ & \Rightarrow & y-4=\frac{4}{5}\left(x-2\right)\\ & \Rightarrow & 5y-20=4x-8\\ & \Rightarrow & 4x-5y-8+20=0\\ & \Rightarrow & 4x-5y+12=0\end{array}$

Therefore, the required equation of the line is $\begin{array}{rcl}4x-5y+12& =& 0\end{array}$