Question

The vertices of a triangle Pqr are P(-3,2) (-1,-4) and R(5,2).
If x and y are the midpoints of PQ and PR Respectively.
shows that 2xy
= QR​

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{The vertices of triangle PQR are P(-3,2), Q(-1-4) and R(5,2)}$

$\textsf{x and y are the midpoints of PQ and PR}$

$\underline{\textbf{To prove:}}$

$\mathsf{2\;XY=QR}$

$\underline{\textbf{Solution:}}$

$\underline{\textsf{Concept used:}}$

$\mathsf{The\;distance\;between\;(x_1,y_1)\;and\;(x_2,y_2)\;is}$

$\boxed{\mathsf{d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}}$

$\textsf{X = Midpoint of PR}$

$\mathsf{X=\left(\dfrac{-3-1}{2},\dfrac{2-4}{2}\right)}$

$\mathsf{X=\left(\dfrac{-4}{2},\dfrac{-2}{2}\right)}$

$\implies\mathsf{X=(-2,-1)}$

$\textsf{Y = Midpoint of PR}$

$\mathsf{Y=\left(\dfrac{-3+5}{2},\dfrac{2+2}{2}\right)}$

$\mathsf{Y=\left(\dfrac{2}{2},\dfrac{4}{2}\right)}$

$\implies\mathsf{Y=(1,2)}$

$\mathsf{QR=\sqrt{(-1-5)^2+(-4-2)^2}}$

$\mathsf{QR=\sqrt{(-6)^2+(-6)^2}}$

$\mathsf{QR=\sqrt{36+36}}$

$\mathsf{QR=\sqrt{72}}$

$\mathsf{QR=2\sqrt{18}}$

$\mathsf{XY=\sqrt{(-2-1)^2+(-1-2)^2}}$

$\mathsf{XY=\sqrt{(-3)^2+(-3)^2}}$

$\mathsf{XY=\sqrt{9+9}}$

$\mathsf{XY=\sqrt{18}}$

$\implies\boxed{\mathsf{2\,XY=QR}}$

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