Question

16
5. Prove that the points (7, 10), (4, 5) and
(10, 15) are the vertices of an isosceles triangle.​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Let the vertices of an isosceles right triangle be A(7,10), B(-2,5) and C(3,-4)

So by distance formula we have,

Distance between two points = (x2−x1)2+(y2−y1)2

AB=(−2−7)2+(5−10)2=81+25=106

BC=(3+2)2+(−4−5)2=(25+81)=106

AC=(3−7)2+(−4−10)2=−16+196=212

∴AB=BC⇒ This implies that ABC is an isosceles triangle.

Also, 

AB2+BC2=106+106=212

∴AB2+BC2=AC2 (Pythagoras theorem)

Hence, proved

∴ Δ ABC is a right triangle (proved)