SATHEE: Straight Line Question 136

Straight Line Question 136

Question: The reflection of the point (4, -13) in the line $ 5x+y+6=0 $ is [EAMCET 1994]

Options:

A) $ (-1,-14) $

B) (3 ,4)

C) (1, 2)

D) (- 4, 13)

Show Answer

Answer:

Correct Answer: A

Solution:

Let $ Q(a,b) $ be the reflection of $ P(4,-13) $ in the line $ 5x+y+6=0 $ . Then the mid-point $ R( \frac{a+4}{2},\frac{b-13}{2} ) $ lies on $ 5x+y+6=0 $ . \ $ 5( \frac{a+4}{2} )+\frac{b-13}{2}+6=0\Rightarrow 5a+b+19=0 $ ……(i) Also $ PQ $ is perpendicular to $ 5x+y+6=0 $ . Therefore $ \frac{b+13}{a-4}\times ( -\frac{5}{1} )=-1\Rightarrow a-5b-69=0 $ …..(ii) Solving (i) and (ii), we get $ a=-1,b=-14 $ .

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Straight Line Question 136

Question: The reflection of the point (4, -13) in the line $ 5x+y+6=0 $ is [EAMCET 1994]

Options:

Answer:

Solution:

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