If the mid-points of the sides BC,CA and AB o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the mid-points be $P(2,1)$ on $BC$,$Q(-1,-2)$ on $CA$,and $R(3,3)$ on $AB$.Since $RQ$ is parallel to $BC$ and $RQ = \frac{1}{2} BC$,the side $

Vedclass · Accepted Answer

$5x-4y=6$

Vedclass · Answer

(B) Let the mid-points be $P(2,1)$ on $BC$,$Q(-1,-2)$ on $CA$,and $R(3,3)$ on $AB$.Since $RQ$ is parallel to $BC$ and $RQ = \frac{1}{2} BC$,the side $BC$ is parallel to the line segment $RQ$.The slope of $RQ$ is $m = \frac{3 - (-2)}{3 - (-1)} = \frac{5}{4}$.Since $BC$ is parallel to $RQ$,the slope of $BC$ is also $m = \frac{5}{4}$.The side $BC$ passes through the point $P(2,1)$.Using the point-slope form,the equation of $BC$ is:$y - 1 = \frac{5}{4}(x - 2)$$4(y - 1) = 5(x - 2)$$4y - 4 = 5x - 10$$5x - 4y = 6$

If the mid-points of the sides $BC$,$CA$ and $AB$ of a triangle $ABC$ are respectively $(2,1)$,$(-1,-2)$ and $(3,3)$,then the equation of the side $BC$ is

Similar Questions

If the extremities of the base of an isosceles triangle are the points $(2a, 0)$ and $(0, a)$ and the equation of one of the sides is $x = 2a$,then the area of the triangle is

If $x_1, x_2, x_3$ and $y_1, y_2, y_3$ are in $G.P.$ with the same common ratio,then the points $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ are:

If the straight lines $2x + 3y - 1 = 0$,$x + 2y - 1 = 0$,and $ax + by - 1 = 0$ form a triangle with the origin as the orthocentre,then $(a, b)$ is equal to

The area (in square units) of the triangle formed by the lines $x=0, y=0$ and $3x+4y=12$ is:

The points $A(2, 1)$,$B(3, -2)$ and $C(a, b)$ are vertices of the rectangle $ABCD$. If the point $P(3, 4)$ lies on the line $CD$,then $5a + 10b = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module