Question

Find the equation of the median from vertex A in triangle ABC, if the coordinates of the vertices are A(-7,3), B(-2,-3), and C(4,2)

Answer 1

The median from A passes through midpoint of BC. Let M be the midpoint of BC whose coordinates, so, will be,

$\longrightarrow\left(x_M,\ y_M\right)=\left(\dfrac{-2+4}{2},\ \dfrac{-3+2}{2}\right)$

$\longrightarrow\left(x_M,\ y_M\right)=\left(1,\ -\dfrac{1}{2}\right)$

Now by two - point form, the equation of the median will be,

$\longrightarrow\dfrac{y-3}{x+7}=\dfrac{-\dfrac{1}{2}-3}{1+7}$

$\longrightarrow\dfrac{y-3}{x+7}=-\dfrac{7}{16}$

$\longrightarrow48-16y=7x+49$

$\longrightarrow\underline{\underline{7x+16y+1=0}}$

Answer 2

Answer:

The equation is 7x + 16y + 1 = 0.

Question :-

• Find the equation of the median from vertex A in triangle ABC, if the coordinates of the vertices are A(-7,3), B(-2,-3), and C(4,2)

Solution :-

$\implies \sf \bigg(x_M,\ y_M\bigg)=\bigg(\dfrac{-2+4}{2},\ \dfrac{-3+2}{2}\bigg)$

$\implies \sf \bigg(x_M,\ y_M\bigg)=\bigg(1,\ -\dfrac{1}{2}\bigg)$

Now by taking two point form, the equation of the median will be,

$\implies \sf \dfrac{y-3}{x+7}=\dfrac{-\dfrac{1}{2}-3}{1+7}$

$\implies \sf \dfrac{y-3}{x+7}=-\dfrac{7}{16}$

$\implies \sf 48-16y=7x+49$

$\implies \sf 7x+16y+1=0$

Therefore, the equation is 7x + 16y + 1 = 0.