Question

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

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

Answer 1

Answer: (B)

$x + y = 5$

Answer 2

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

So, $3 a + 2 b = 13$ .....(i)

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

So, $4 b - a = 5$ .....(ii)

By solving (i) and (ii), $a = 3 , b = 2$

$P ( a , b ) \rightarrow ( 3,2 )$and $Q ( b , a ) \rightarrow ( 2,3 )$

Now, equation of PQ

$y - y _ { 1 } = \frac { y _ { 2 } - y _ { 1 } } { x _ { 2 } - x _ { 1 } } \left( x - x _ { 1 } \right) \Rightarrow y - 2 = \frac { 3 - 2 } { 2 - 3 } ( x - 3 )$

⇒ $y - 2 = - ( x - 3 ) \Rightarrow x + y = 5$.