Question

The equation of line passing through the point of intersection of lines $3x-2y-1=0$ and $x-4y+3=0$ and the point $\left(\mathrm{\pi },0\right)$ is:

• A. $x-y=\mathrm{\pi }$
• B. $x-y=\mathrm{\pi }(\mathrm{y}+1)$
• C. $x-y=\mathrm{\pi }(1-y)$
• D. $x+y=\mathrm{\pi }(1-y)$

Answer 1

The correct option is C

$x-y=\mathrm{\pi }(1-y)$

Explanation for the correct option:

Finding the equation of line,

Let the required line be $AB$.

Calculating the point of intersection of $3x-2y-1=0\dots \dots \left(1\right)$ and $x-4y+3=0$:

Multiplying $x-4y+3=0$ by $3$, we get

$3x-12y+9=0\dots \dots \left(2\right)$

Subtracting equations $\left(1\right)$ and $\left(2\right)$, we get

$\begin{array}{rcl}3x-2y-1-\left(3x-12y+9\right)& =& 0\\ & \Rightarrow & -2y-10=0\\ & \Rightarrow & 10y=10\\ & \Rightarrow & y=1\end{array}$

Substituting the value of $y$in equation $\left(1\right)$ we get

$$\begin{gathered} 3x-2y-1=0 \\ 3x-2\left(1\right)-1=0 \\ 3x-3=0 \\ x=1 \end{gathered}$$

Hence, the point of intersection is $(1,1)$.

Using the formula for the line passing through points, to find the equation of line $AB$

$\begin{array}{rcl}& \Rightarrow & y-1=\frac{1-0}{1-\mathrm{\pi }}(x-1)\\ & \Rightarrow & (y-1)(1-\mathrm{\pi })=\mathrm{x}-1\\ & \Rightarrow & x-y=\mathrm{\pi }(1-\mathrm{y})\end{array}$

Hence, the correct answer is Option (C).