Question

If 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 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 C

$x+y=-5$

Explanation for Correct Option:

Step 1: Solving given equation

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

$\Rightarrow 3a+2b=13$ ..(i)

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

$\Rightarrow 4b-a=5$ ..(ii)

Step 2: Solving (i) and (ii), we get

$a=3$ and $b=2$

So$\mathrm{P}=(3,2)$ and $\mathrm{Q}=(2,3)$

Here $\mathrm{m}=\frac{(3–2)}{(2–3)}=-1$

Step 3: Finding equation of line

Equation of line is : $y–y_1=m(x–x_1)$

$\therefore (y–2)=-1(x–3)$

$\therefore x+y=-5$

Hence option (C) is the Correct.