SATHEE: Straight Line Question 143

Straight Line Question 143

Question: If each of the points, $ (x_1,4) $ , $ (-2,y_1) $ lies on the line joining the points (2, -1) and (5, -3), then the point $ p(x_1,y_1) $ lies on the line

Options:

A) $ 6(x+y)-25=0 $

B) $ 2x+6y+1=0 $

C) $ 2x+3y-6=0 $

D) $ 6(x+y)+25=0 $

Show Answer

Answer:

Correct Answer: B

Solution:

[b] The equation of the line joining the points (2,-1) and (5,-3) is given by $ y+1=\frac{-1+3}{2-5}(x-2) $ Or $ 2x+3y-1=0 $ Since $ (x_1,4) $ and $ (-2,y_1) $ lie on $ 2x+3y-1=0 $ , we have $ 2x_1+12-1=0 $ or $ x_1=-\frac{11}{2} $ And $ -4+3y_1-1=0 $ or $ y_1=\frac{5}{3} $ Thus, $ (x_1,y_1) $ satisfies $ 2x+6y+1=0 $ .

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Straight Line Question 143

Question: If each of the points, $ (x_1,4) $ , $ (-2,y_1) $ lies on the line joining the points (2, -1) and (5, -3), then the point $ p(x_1,y_1) $ lies on the line

Options:

Answer:

Solution:

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