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

# 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

1. A

x + y = 7

2. B

x + y = 9

3. C

x + y = 2

4. D

x + y = 21

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### Solution:

$\begin{array}{l}Slope\text{\hspace{0.17em}\hspace{0.17em}}of\text{\hspace{0.17em}\hspace{0.17em}}PQ\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{\frac{31}{5}-\frac{14}{5}}{\frac{14}{5}-\frac{31}{5}}=\frac{17}{-17}=-1\\ Equation\text{\hspace{0.17em}\hspace{0.17em}}of\text{\hspace{0.17em}\hspace{0.17em}}\overline{PQ}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}y=\frac{14}{5}=-1\left(x-\frac{31}{5}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}5y-14=-5x+31\\ ⇒\text{\hspace{0.17em}}5x+5y-45=0\\ ⇒\text{\hspace{0.17em}}x+y=9\end{array}$

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