Question

If point $P(a,b)$ lies on the straight line $3x+2y-16=0$ and the point $Q(b,a)$ lies on the straight line $4x-y-5=0$. The equation of the line $PQ$ is

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

Answer 1

The correct option is B $x+y=5$

Since $P(a,b)$ lies on $3x+2y=13$, we get

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

Similarly, since $Q(b,a)$ lies on $4x-y=5$, we get

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

Solve $\left(1\right),\left(2\right)$ using substitution method i.e. substituting $a=4b-5$ in $\left(1\right)$ gives $b=2$ and $a=4b-5=3$

Equation of line passing through two points $(x_1,y_1),(x_2,y_2)$ is $\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$

$(x_1,y_1)=(a,b),(x_2,y_2)=(b,a)\Rightarrow \overline{PQ}=\frac{y-b}{a-b}=\frac{x-a}{b-a}\Rightarrow y-b=-(x-a)\Rightarrow x+y=a+b=5$