Question

show that the points (7,9),(3,-7) and (-3,3) are the vertices of a right angled isosceles triangle​

Answer 1

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let the points are...A (7,9);B(3,-7);C(-3,3).

now,the points would be the vertices of a right angled triangle.. if ..any one side is equal to the root of sum of other two sides...

$\underline{\large\mathcal\red{!! solution!!}$

$ab = \sqrt{(7 - 3) {}^{2} + (9 + 7) {}^{2} } = \sqrt{4 {}^{2} + 16 {}^{2} } = 4 \sqrt{17} \\ \\ bc = \sqrt{(3 + 3) {}^{2} + ( - 7 - 3) {}^{2} } = \sqrt{6 {}^{2} + 10 {}^{2}} = 2 \sqrt{34} \\ \\ ca = \sqrt{( - 3 - 7) {}^{2} + (3 - 9) {}^{2} } = \sqrt{10 {}^{2} + 6 {}^{2} } = 2 \sqrt{34}$

Now ,from Pythagorean theorem....for a right angled triangle...

$bc {}^{2} + ca {}^{2} = (2 \sqrt{34} ) {}^{2} + (2 \sqrt{34} ) {}^{2} = 272 \\ = (4 \sqrt{17} ) {}^{2} = ab {}^{2}$

so, ∆ABC is a right angled triangle.....

therefore,A,B,C are the vertices of a Right angled triangle.(proved).

Answer 2

Answer:

Hey mate plzz refer to the attachment