SATHEE: Straight Line Question 346

Straight Line Question 346

Question: The point $ P(a,\ 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 line PQ is [MP PET 1999]

Options:

A) $ x-y=5 $

B) $ x+y=5 $

C) $ x+y=-\ 5 $

D) $ x-y=-\ 5 $

Show Answer

Answer:

Correct Answer: B

Solution:

Point $ P(a,b) $ is on $ 3x+2y=13 $ So, $ 3a+2b=13 $ …..(i) Point $ Q(b,a) $ is on $ 4x-y=5 $ So, $ 4b-a=5 $ …..(ii) By solving (i) and (ii), $ a=3,b=2 $ $ P(a,b)\to (3,2) $ and $ Q(b,a)\to (2,3) $ Now, equation of PQ $ y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\Rightarrow y-2=\frac{3-2}{2-3}(x-3) $
Þ $ y-2=-(x-3)\Rightarrow x+y=5 $ .

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Straight Line Question 346

Question: The point $ P(a,\ 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 line PQ is [MP PET 1999]

Options:

Answer:

Solution:

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