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A(8, 6), B(10, -10) and C(4, -4) are the vertices of triangle ABC. If P is the mid-point of side AB and Q is mid-point of side AC, show that PQ is parallel to side BC also show 2 × PQ = BC.

Section Formula

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Answer

By formula,

Mid-point = (x1+x22,y1+y22)\Big(\dfrac{x1 + x2}{2}, \dfrac{y1 + y2}{2}\Big)

Given, P is mid-point of AB.

∴P=(8+102,6+(−10)2)⇒P=(182,−42)⇒P=(9,−2).\therefore P = \Big(\dfrac{8 + 10}{2}, \dfrac{6 + (-10)}{2}\Big) \\[1em] \Rightarrow P = \Big(\dfrac{18}{2}, \dfrac{-4}{2}\Big) \\[1em] \Rightarrow P = (9, -2).

Given, Q is mid-point of AC.

∴Q=(8+42,6+(−4)2)⇒Q=(122,22)⇒Q=(6,1).\therefore Q = \Big(\dfrac{8 + 4}{2}, \dfrac{6 + (-4)}{2}\Big) \\[1em] \Rightarrow Q = \Big(\dfrac{12}{2}, \dfrac{2}{2}\Big) \\[1em] \Rightarrow Q = (6, 1).

By formula,

Slope = y2−y1x2−x1\dfrac{y2 - y1}{x2 - x1}

Slope of PQ =1−(−2)6−9=3−3=−1.Slope of BC =−4−(−10)4−10=−4+10−6=6−6=−1.\text{Slope of PQ }= \dfrac{1 - (-2)}{6 - 9} \\[1em] = \dfrac{3}{-3} \\[1em] = -1. \\[1em] \text{Slope of BC }= \dfrac{-4 - (-10)}{4 - 10} \\[1em] = \dfrac{-4 + 10}{-6} \\[1em] = \dfrac{6}{-6} \\[1em] = -1.

Since, slope of PQ = slope of BC = -1

∴ PQ || BC.

By distance formula,

Distance between two points = (y2−y1)2+(x2−x1)2\sqrt{(y2 - y1)^2 + (x2 - x1)^2}

PQ=[1−(−2])2+(6−9)2=[3]2+(−3)2=9+9=18=32 cm.BC=[−4−(−10)]2+(4−10)2=[−4+10]2+(−6)2=62+(−6)2=36+36=72=62=2×32=2×PQ.PQ = \sqrt{[1 - (-2])^2 + (6 - 9)^2} \\[1em] = \sqrt{[3]^2 + (-3)^2} \\[1em] = \sqrt{9 + 9} \\[1em] = \sqrt{18} \\[1em] = 3\sqrt{2} \text{ cm}. \\[1em] BC = \sqrt{[-4 - (-10)]^2 + (4 - 10)^2} \\[1em] = \sqrt{[-4 + 10]^2 + (-6)^2} \\[1em] = \sqrt{6^2 + (-6)^2} \\[1em] = \sqrt{36 + 36} \\[1em] = \sqrt{72} \\[1em] = 6\sqrt{2} \\[1em] = 2 \times 3\sqrt{2} \\[1em] = 2 \times PQ.

Hence, proved that PQ || BC and 2 × PQ = BC.

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