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Show that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right triangle.

Coordinate Geometry

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Answer

Let points be A(7, 10), B(-2, 5) and C(3, -4).

Show that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right triangle. Coordinate Geometry, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

By distance formula,

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

AB=(−2−7)2+(5−10)2=(−9)2+(−5)2=81+25=106 units.BC=[3−(−2)]2+[−4−5]2=52+(−9)2=25+81=106 units.AC=(3−7)2+(−4−10)2=(−4)2+(−14)2=16+196=212 units.AB = \sqrt{(-2 - 7)^2 + (5 - 10)^2} \\[1em] = \sqrt{(-9)^2 + (-5)^2} \\[1em] = \sqrt{81 + 25} \\[1em] = \sqrt{106} \text{ units}. \\[1em] BC = \sqrt{[3 - (-2)]^2 + [-4 - 5]^2} \\[1em] = \sqrt{5^2 + (-9)^2} \\[1em] = \sqrt{25 + 81} \\[1em] = \sqrt{106} \text{ units}. \\[1em] AC = \sqrt{(3 - 7)^2 + (-4 - 10)^2} \\[1em] = \sqrt{(-4)^2 + (-14)^2} \\[1em] = \sqrt{16 + 196} \\[1em] = \sqrt{212} \text{ units}.

⇒ AB2 + BC2 = (106)2+(106)2(\sqrt{106})^2 + (\sqrt{106})^2

= 106 + 106

= 212

= AC2.

Since, AB = BC and AB2 + BC2 = AC2.

∴ ABC is an isosceles right angle triangle, right angled at B.

Hence, proved that (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles triangle.

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