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The points A(0, 3), B(-2, a) and C(-1, 4) are the vertices of a right angled triangle at A, find the value of a.

Coordinate Geometry

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Answer

By distance formula,

The points A(0, 3), B(-2, a) and C(-1, 4) are the vertices of a right angled triangle at A, find the value of a. Coordinate Geometry, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

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

AB=(−2−0)2+(a−3)2=(−2)2+a2+9−6a=4+a2+9−6a=a2−6a+13.BC=[−1−(−2)]2+(4−a)2=(−1+2)2+(4−a)2=(1)2+16+a2−8a=1+16+a2−8a=a2−8a+17.AC=(−1−0)2+(4−3)2=(−1)2+(1)2=1+1=2.AB = \sqrt{(-2 - 0)^2 + (a - 3)^2} \\[1em] = \sqrt{(-2)^2 + a^2 + 9 - 6a} \\[1em] = \sqrt{4 + a^2 + 9 - 6a} \\[1em] = \sqrt{a^2 - 6a + 13}. \\[1em] BC = \sqrt{[-1 - (-2)]^2 + (4 - a)^2} \\[1em] = \sqrt{(-1 + 2)^2 + (4 - a)^2} \\[1em] = \sqrt{(1)^2 + 16 + a^2 - 8a} \\[1em] = \sqrt{1 + 16 + a^2 - 8a} \\[1em] = \sqrt{a^2 - 8a + 17}. \\[1em] AC = \sqrt{(-1 - 0)^2 + (4 - 3)^2} \\[1em] = \sqrt{(-1)^2 + (1)^2} \\[1em] = \sqrt{1 + 1} \\[1em] = \sqrt{2}.

By pythagoras theorem,

AB2 + AC2 = BC2

⇒(a2−6a+13)2+(2)2=(a2−8a+17)2⇒a2−6a+13+2=a2−8a+17⇒a2−a2−6a+8a=17−15⇒2a=2⇒a=1.\Rightarrow \Big(\sqrt{a^2 - 6a + 13}\Big)^2 + \Big(\sqrt{2}\Big)^2 = \Big(\sqrt{a^2 - 8a + 17}\Big)^2 \\[1em] \Rightarrow a^2 - 6a + 13 + 2 = a^2 - 8a + 17 \\[1em] \Rightarrow a^2 - a^2 - 6a + 8a = 17 - 15 \\[1em] \Rightarrow 2a = 2 \\[1em] \Rightarrow a = 1.

Hence, value of a = 1.

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