Question

The point $\mathrm{P}(\mathrm{a},\mathrm{b})$ lies on the straight line $3x+2y=13$ and the point $Q(b,a)$ lies on the straight line $4x-y=5$, then the equation of line $\mathrm{PQ}$ is

• A. $x-y=5$
• B. $x+y=5$
• C. $x+y=-5$
• D. $x-y=-5$

Answer 1

The correct option is B

$x+y=5$

Compute the required equation:

Given :

$\mathrm{P}(\mathrm{a},\mathrm{b})$ lies on the straight line $3x+2y=13$

$\Rightarrow 3a+2b=13$…………$\left(1\right)$

$Q(b,a)$ lies on the straight line $4x-y=5$,

$\Rightarrow 4b-a=5$…………….$\left(2\right)$

Multiply equation $\left(1\right)$ by $2$ and subtract equation $\left(2\right)$ from $\left(1\right)$

$6a+4b-4b+a=26-5$

$\Rightarrow$ $7a=21$

$\Rightarrow$ $a=3$

Substitute the value of $a$ in equation $\left(1\right)$

$$\begin{gathered} 3\left(3\right)+2b=13 \\ \Rightarrow 2b=13-9 \\ \Rightarrow 2b=4 \\ \Rightarrow b=2 \end{gathered}$$

Therefore, $\mathrm{P}$ is $(3,2)$ and $\mathrm{Q}$ is $\left(2,3\right)$.

Since the equation of the line is $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

Therefore the equation of the line $\mathrm{PQ}$ is,

$y-2=\frac{3-2}{2-3}(x-3)$

$\Rightarrow$ $y=2-(x-3)$

$\Rightarrow$ $y=-x+5$

$\Rightarrow x+y=5$

Hence option $\left(B\right)$ is the correct option.