Question

P($-$3, 4), Q(2, 3), R($-$2, $-$5) are the vertices of the $∆$PQR. Find the equations of all the medians of $∆$PQR.

Answer 1

The given points P($-$3, 4), Q(2, 3) and R($-$2, -5) are the vertices of $△$PQR and
PB, RA and QC are the medians of the triangle, i.e., A, B and C are the midpoints of sides PQ, QR and PR.

Now, we have:

x- coordinate of A = $\frac{-3+2}{2}=\frac{-1}{2}$

y- coordinate of A = $\frac{3+4}{2}=\frac{7}{2}$

∴ The coordinates of point A are $\left(\frac{-1}{2},\frac{7}{2}\right)$.
Now, slope of the median RA is given by
$$\begin{gathered} m=\frac{y_1-y_2}{x_1-x_2} \\ =\frac{-5-{\displaystyle \frac{7}{2}}}{-2-\left({\displaystyle \frac{-1}{2}}\right)} \\ =\frac{17}{3} \end{gathered}$$

Also, equation of the median RA is given by
y $-$ y₁ = m (x $-$ x₁)
⇒ y $-$ ($-$5)= $\frac{17}{3}$ {x $-$ ($-$2)}
⇒ 3y + 15= 17x +34
⇒ 17x $-$ 3y +19 = 0

Now, we have:
x- coordinate of B = $\frac{-2+2}{2}=0$

y- coordinate of B = $\frac{3-5}{2}=-1$

∴ The coordinates of point B are (0,$-$1).
Now, slope of the median PB is given by

$$\begin{gathered} m=\frac{y_1-y_2}{x_1-x_2} \\ =\frac{4-(-1)}{-3-0} \\ =\frac{-5}{3} \end{gathered}$$

Also, equation of the median PB is given by
y $-$ y₁ = m (x $-$ x₁)
⇒ y $-$4 = $\frac{-5}{3}$ {x $-$ ($-$3)}
⇒ 3y $-$12= $-$5x $-$15
⇒ 5x + 3y + 3= 0

Now, we have:
x-coordinate of C= $\frac{-3-2}{2}=\frac{-5}{2}$

y-coordinate of C = $\frac{-5+4}{2}=\frac{-1}{2}$

∴ The coordinates of point C are $\left(\frac{-5}{2},\frac{-1}{2}\right)$.
Now, slope of the median QC is given by

$$\begin{gathered} m=\frac{y_1-y_2}{x_1-x_2} \\ =\frac{3-\left({\displaystyle \frac{-1}{2}}\right)}{2-\left({\displaystyle \frac{-5}{2}}\right)} \\ =\frac{7}{9} \end{gathered}$$

Also, equation of the median QC is given by
y $-$ y₁ = m (x $-$ x₁)
⇒ y $-$3 = $\frac{7}{9}$ (x $-$ 2)
⇒ 9y $-$27= 7x $-$14
⇒ 7x $-$ 9y +13 = 0