If coordinates of vertices of triangle ABC be (-1, 6) , (-3, -9) , and (5, -8) respectively, then eq

English

Gujarati

Hindi

9.Straight Line

easy

If the coordinates of the vertices of the triangle $ABC$ be $(-1, 6)$, $(-3, -9)$, and $(5, -8)$ respectively, then the equation of the median through $C$ is

A

$13x - 14y - 47 = 0$

B

$13x - 14y + 47 = 0$

C

$13x + 14y + 47 = 0$

D

$13x + 14y - 47 = 0$

Solution

(c) Required equation of median is $y + 8 = \frac{{ – \frac{3}{2} + 8}}{{ – 2 – 5}}(x – 5)$

$ \Rightarrow 13x + 14y + 47 = 0$.

Standard 11

Mathematics

Share

Similar Questions

The base of an equilateral triangle with side $2 a$ lies along the $y$ -axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

medium

If the straight line $ax + by + c = 0$ always passes through $(1, -2),$ then $a, b, c$ are

easy

The opposite angular points of a square are $(3,\;4)$ and $(1,\; – \;1)$. Then the co-ordinates of other two points are

medium

$P (x, y)$ moves such that the area of the triangle formed by $P, Q (a , 2 a)$ and $R (- a, – 2 a)$ is equal to the area of the triangle formed by $P, S (a, 2 a)\,\,\, \&\,\, \,T (2 a, 3 a)$. The locus of $'P'$ is a straight line given by :

normal

The equation of the line which makes right angled triangle with axes whose area is $6$ sq. units and whose hypotenuse is of $5$ units, is

medium