SATHEE: Straight Line Question 135

Straight Line Question 135

Question: The image of a point $ A(3,8) $ in the line $ x+3y-7=0 $ , is [RPET 1991]

Options:

A) $ (-1,-4) $

B) $ (-3,-8) $

C) $ (1,-4) $

D) $ (3,8) $

Show Answer

Answer:

Correct Answer: A

Solution:

Equation of the line passing through (3, 8) and perpendicular to $ x+3y-7=0 $ is $ 3x-y-1=0 $ . The intersection point of both the lines is (1, 2). Now let the image of $ A(3,8) $ be $ {A}’(x_1,y_1), $ then point (1, 2) will be the midpoint of $ A{A}’ $ .
Þ $ \frac{x_1+3}{2}=1\Rightarrow x_1=-1 $ and $ \frac{y_1+8}{2}=2 $ Þ $ y_1=-4 $ . Hence the image is (?1, ?4).

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Straight Line Question 135

Question: The image of a point $ A(3,8) $ in the line $ x+3y-7=0 $ , is [RPET 1991]

Options:

Answer:

Solution:

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