Let PS be the median of the triangle with vertices P (2, 2), Q (6-1) and R (7, 3). The equation of the line passing through (1,–1) and parallel to PS is:
(A) 2x-9 y-7=0 (B) 2 x-9 y-11=0
(C) 2 x+9 y-11=0 (D) 2 x-9 y-11=0

Question

AskIITian Expert Priyasheel - IITD · Answer

Coordinates of mid point of Q and R which is S, will be (13/2, 1)

Equation of line passing through P and S will be

(y-2)/(x-2) = (2-1)/(2-13/2), OR 2x+9y -22=0,

Equation of a line parallel to 2x+9y -22=0 will be 2x+9y + k=0,

This line passes through (1,-1), so 2*1+9*-1+k=0 => k=7.

So the line is 2x+9y +7=0

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Let PS be the median of the triangle with vertices P (2, 2), Q (6-1) and R (7, 3). The equation of the line passing through (1,–1) and parallel to PS is: (A) 2x-9 y-7=0 (B) 2 x-9 y-11=0 (C) 2 x+9 y-11=0 (D) 2 x-9 y-11=0

Let PS be the median of the triangle with vertices P (2, 2), Q (6-1) and R (7, 3). The equation of the line passing through (1,–1) and parallel to PS is:
(A) 2x-9 y-7=0 (B) 2 x-9 y-11=0
(C) 2 x+9 y-11=0 (D) 2 x-9 y-11=0

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Let PS be the median of the triangle with vertices P (2, 2), Q (6-1) and R (7, 3). The equation of the line passing through (1,–1) and parallel to PS is: (A) 2x-9 y-7=0 (B) 2 x-9 y-11=0 (C) 2 x+9 y-11=0 (D) 2 x-9 y-11=0

Let PS be the median of the triangle with vertices P (2, 2), Q (6-1) and R (7, 3). The equation of the line passing through (1,–1) and parallel to PS is: (A) 2x-9 y-7=0 (B) 2 x-9 y-11=0 (C) 2 x+9 y-11=0 (D) 2 x-9 y-11=0

1 Answers

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Let PS be the median of the triangle with vertices P (2, 2), Q (6-1) and R (7, 3). The equation of the line passing through (1,–1) and parallel to PS is:
(A) 2x-9 y-7=0 (B) 2 x-9 y-11=0
(C) 2 x+9 y-11=0 (D) 2 x-9 y-11=0