The point P (a,b) lies on the straight line 3x-2y=13 and the point Q (b, a) lies on the staright line 4x-y=5 then the equation of line PQ is - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
x + y = 7
B
x + y = 9
C
x + y = 2
D
x + y = 21

Solution:

$\begin{array}{l} P (a, b) l i e s o n 3 x - 2 y - 13 = 0 \Rightarrow 3 a - 2 b - 13 = 0 - - (1) \\ Q (b, a) l i e s o n 4 x - y - 5 = 0 \Rightarrow 4 b - a - 5 = 0 \Rightarrow a - 4 b + 5 = 0 - - (2) \\ S o l v i n g (1) & (2) \\ a b 1 \\ \begin{matrix} - 2 & - 13 & 3 & - 2 \\ - 4 & 5 & 1 & - 4 \end{matrix} \\ \Rightarrow \frac{a}{- 10 - 52} = \frac{b}{- 13 - 15} = \frac{1}{- 12 + 2} \\ \Rightarrow \frac{a}{- 62} = \frac{b}{- 28} = \frac{1}{- 10} \\ ∴ a = \frac{- 62}{- 10} = \frac{31}{5}, b = \frac{- 28}{- 10} = \frac{14}{5} \\ ∴ P (\frac{31}{5}, \frac{14}{5}) a n d Q (\frac{14}{5}, \frac{31}{5}) \end{array}$

$\begin{array}{l} S l o p e o f P Q = \frac{\frac{31}{5} - \frac{14}{5}}{\frac{14}{5} - \frac{31}{5}} = \frac{17}{- 17} = - 1 \\ E q u a t i o n o f \bar{P Q} y = \frac{14}{5} = - 1 (x - \frac{31}{5}) \\ \Rightarrow 5 y - 14 = - 5 x + 31 \\ \Rightarrow 5 x + 5 y - 45 = 0 \\ \Rightarrow x + y = 9 \end{array}$

Solution:

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