P(3, 4), Q(7, -2) and R(-2, -1) are the vertices of triangle PQR. Write down the equation of the median of the triangle through R.

$x - 7 y - 3 = 0$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$2 x - 7 y + 3 = 0$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$2 x - 7 y - 3 = 0$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$2 x + 7 y - 3 = 0$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
$2 x - 7 y - 3 = 0$

The vertices of $Δ P Q R$ are P(3, 4), Q(7, -2) and R(-2, -1)
Let S be the mid-point of PQ, then RS is the median through R

$∴$ Coordinates of S are
$(\frac{3 + 7}{2}, \frac{4 - 2}{2}) = (5, 1)$
Slope of RS, $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{1 - (- 1)}{5 - (- 2)} = \frac{2}{7}$
$∴$ The equation of the median RS is
$y - y_{1} = m (x - x_{1})$
$\Rightarrow y - (- 1) = \frac{2}{7} (x + 2)$
$\Rightarrow y + 1 = \frac{2}{7} (x + 2)$
$\Rightarrow 7 y + 7 = 2 x + 4$
$\Rightarrow 2 x - 7 y - 3 = 0$
Hence, the equation of the median of the triangle through R is $2 x - 7 y - 3 = 0$

Suggest Corrections

Similar questions

Q. Let

P S

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

P S

Q. The vertices of a triangle are

A (- 1, - 7), B (5, 1),

and

C (1, 4) .

The equation of the bisector of

∠ A B C

is ___

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

Q. The equation of common tangent(s) to the hyperbola

9 x^{2} - 16 y^{2} = 144

and circle

x^{2} + y^{2} = 9

is/are

Q. The equation of the plane equally inclined to the lines

L_{1} : \frac{x - 1}{2} = \frac{y}{- 2} = \frac{z + 2}{- 1}

and

L_{2} : \frac{x + 3}{8} = \frac{y - 4}{1} = \frac{z}{- 4}

and passing through the origin is/are

Join BYJU'S Learning Program