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If A(-3, 2), B(α, β) and C(-1, 4) are the vertices of an isosceles triangle, prove that α + β = 1, given AB = BC.

Coordinate Geometry

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Answer

Given,

A(-3, 2), B(α, β) and C(-1, 4) are the vertices of an isosceles triangle

By distance formula,

If A(-3, 2), B(α, β) and C(-1, 4) are the vertices of an isosceles triangle, prove that α + β = 1, given AB = BC. Coordinate Geometry, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

d=(x2−x1)2+(y2−y1)2d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2}

As, AB = BC

∴[α−(−3)]2+(β−2)2=(−1−α)2+(4−β)2\therefore \sqrt{[α - (-3)]^2 + (β - 2)^2} = \sqrt{(-1 - α)^2 + (4 - β)^2}

On squaring both sides,

⇒[α−(−3)]2+(β−2)2=(−1−α)2+(4−β)2⇒[α+3]2+(β−2)2=(−1−α)2+(4−β)2⇒α2+32+6α+β2+4−4β=1+α2+2α+16+β2−8β⇒α2+9+6α+β2+4−4β=α2+2α+17+β2−8β⇒α2−α2+β2−β2+6α−2α−4β+8β=17−9−4⇒4α+4β=4⇒4(α+β)=4⇒α+β=1.\Rightarrow [α - (-3)]^2 + (β - 2)^2 = (-1 - α)^2 + (4 - β)^2 \\[1em] \Rightarrow [α + 3]^2 + (β - 2)^2 = (-1 - α)^2 + (4 - β)^2 \\[1em] \Rightarrow α^2 + 3^2 + 6α + β^2 + 4 - 4β = 1 + α^2 + 2α + 16 + β^2 - 8β \\[1em] \Rightarrow α^2 + 9 + 6α + β^2 + 4 - 4β = α^2 + 2α + 17 + β^2 - 8β \\[1em] \Rightarrow α^2 - α^2 + β^2 - β^2 + 6α - 2α - 4β + 8β = 17 - 9 - 4 \\[1em] \Rightarrow 4α + 4β = 4 \\[1em] \Rightarrow 4(α + β) = 4 \\[1em] \Rightarrow α + β = 1.

Hence, proved that α + β = 1.

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