Question

show that the triangle whose. vertices (8,-4),(9,5) and (0,4) is an isosceles​

Answer 1

Step-by-step explanation:

A[8,-4],B[9,5],C[0,4]

AB=√

$\sqrt{( x2 - x1)}^{2} + \sqrt{} (y2 - y1) {}^{2}$

$\sqrt{(9 - 8) {}^{2} } + \sqrt{(5 + 4) {}^{2} }$

$\sqrt{(1) { }^{2} } + \sqrt{(9) {}^{2} }$

$\sqrt{1 + 81}$

$= \sqrt{82}$

BC=

$= \sqrt{(0 - 9) {}^{2} } + \sqrt{(4 - 5)} {}^{2}$

$= \sqrt{( - 9) { }^{2} } + \sqrt{( - 1)} {}^{2}$

$= \sqrt{81 + 1}$

$= \sqrt{82}$

CA=

$\sqrt{(8 - 0)} {}^{2} + \sqrt{ (- 4 -4) } {}^{2}$

$\sqrt{(8) { }^{2} } + \sqrt{ (- 8)} {}^{2}$

$\sqrt{64 + 64}$

$= \sqrt{128}$

so it is an isosceles triangle

Answer 2

Step-by-step explanation:

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