If PS is the median of the triangle with vertices P(2,2), Q (6,-1) and R(7,3), then equation of the line passing through (1,-1) and parallel to PS is

Question

If PS is the median of the triangle with vertices P(2,2), Q (6,-1) and R(7,3), then equation of the line passing through (1,-1) and parallel to PS is (A) 4x - 7y - 11 = 0 (B) 2x + 9y + 7 = 0 (C) 4 x + 7 y + 3 = 0 (D) 2x - 9y - 11 = 0

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/d92dcd2a95c68fef550647d2ed533bb1-.png" /> S is mid-point of QR So S=((7+6/2), (3-1/2)) =((13/2), 1) Slope of PS=(2-1/2-(13)2)=-(2/9) Equation of line ⇒ y-(-1)=-(2/9)(x-1) 9 y+9=-2 x+2 ⇒ 2 x+9 y+7=0

Q. If $PS$ is the median of the triangle with vertices $P (2, 2), Q (6, - 1)$ and $R (7, 3)$ , then equation of the line passing through $(1, - 1)$ and parallel to $PS$ is

$4 x - 7 y - 11 = 0$

$2 x + 9 y + 7 = 0$

$4 x + 7 y + 3 = 0$

$2 x - 9 y - 11 = 0$

$S$ is mid-point of $QR$
So $S = (\frac{7 + 6}{2}, \frac{3 - 1}{2})$
$= (\frac{13}{2}, 1)$
Slope of $PS = \frac{2 - 1}{2 - \frac{13}{2}} = - \frac{2}{9}$
Equation of line $\Rightarrow y - (- 1) = - \frac{2}{9} (x - 1)$
$9 y + 9 = - 2 x + 2$
$\Rightarrow 2 x + 9 y + 7 = 0$

Q. If PS is the median of the triangle with vertices P(2,2),Q(6,−1) and R(7,3), then equation of the line passing through (1,−1) and parallel to PS is

4x−7y−11=0

2x+9y+7=0

4x+7y+3=0

2x−9y−11=0