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A(-4, -1), B(-1, 2) and C(α, 5) are the vertices of an isosceles triangle. Find the value of α given that AB is the unequal side.

Coordinate Geometry

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Answer

It is given that

A(-4, -1), B(-1, 2) and C(α, 5) are the vertices of an isosceles triangle and AB is the unequal side.

∴ AC = BC.

A(-4, -1), B(-1, 2) and C(α, 5) are the vertices of an isosceles triangle. Find the value of α given that AB is the unequal side. Coordinate Geometry, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

By distance formula,

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

Since, AC = BC

⇒[α−(−4)]2+[5−(−1)]2=[α−(−1)]2+[5−2]2⇒[α+4]2+[5+1]2=[α+1]2+[3]2⇒α2+16+8α+62=α2+1+2α+9\Rightarrow \sqrt{[α - (-4)]^2 + [5 - (-1)]^2} = \sqrt{[α - (-1)]^2 + [5 - 2]^2} \\[1em] \Rightarrow \sqrt{[α + 4]^2 + [5 + 1]^2} = \sqrt{[α + 1]^2 + [3]^2} \\[1em] \Rightarrow \sqrt{α^2 + 16 + 8α + 6^2} = \sqrt{α^2 + 1 + 2α + 9}

On squaring both sides we get,

⇒α2+16+8α+62=α2+1+2α+9⇒α2+8α+16+36=α2+2α+10⇒α2−α2+8α−2α=10−16−36⇒6α=−42⇒α=−426⇒α=−7.\Rightarrow α^2 + 16 + 8α + 6^2 = α^2 + 1 + 2α + 9 \\[1em] \Rightarrow α^2 + 8α + 16 + 36 = α^2 + 2α + 10 \\[1em] \Rightarrow α^2 - α^2 + 8α - 2α = 10 - 16 - 36 \\[1em] \Rightarrow 6α = -42 \\[1em] \Rightarrow α = -\dfrac{42}{6} \\[1em] \Rightarrow α = -7.

Hence, α = -7.

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