Question

In triangle ABC the co-ordinates of vertices A, B and C are (4, 7), (-2, 3) and (0, 1) respectively. Find the equations of medians passing through vertices A, B and C.

• A. $x-y+3=0$
• B. $x-4y+14=0$
• C. $x+y+3=0$
• D. $4x-y+1=0$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $x-y+3=0$
B $x-4y+14=0$
D $4x-y+1=0$

A median of a triangle is a line segment that joins the vertex of a triangle to the midpoint of the opposite side.
Mid
point of two points ${(x}_1,y_1)$ and ${(x}_2,y_2)$ is calculated by the formula $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
Using this formula,
mid point of AB $=(\frac{4-2}{2},\frac{7+3}{2})=(1,5)$
mid point of BC $=(\frac{-2+0}{2},\frac{3+1}{2})=(-1,2)$
mid point of CA $=(\frac{0+4}{2},\frac{1+7}{2})=(2,4)$
Equation
of a line joining two points ${(x}_1,y_1)$ and ${(x}_2,y_2)$ is given by the formula $y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)(x-x_1)$
Equation of Median passing through
A is the equation passing through A $(4,7)$ and Midpoint of BC $(-1.2)$ is $y-7=\left(\frac{2-7}{-1-4}\right)(x-4)$
$=>y-7=\frac{-5}{-5}(x-4)$
$=>y-7=x-4$
$=>x-y+3=0$
Equation of Median passing through B is the equation passing through B
$(-2,3)$ and Midpoint of AC $(2,4)$ is $y-3=\left(\frac{4-3}{2-(-2)}\right)(x-(-2\left)\right)$
$=>y-3=\frac{1}{4}(x+2)$
$=>4y-12=x+2$
$=>x-4y+14=0$
Equation of Median passing through C is the equation passing through C
$(0,1)$ and Midpoint of AB $(1,5)$ is $y-1=\left(\frac{5-1}{1-0}\right)(x-0)$
$=>y-1=\frac{4}{1}\left(x\right)$
$=>y-1=4x$
$=>4x-y+1=0$