Question

Show that the points A(-5, 6), B(3, 0) and C(9, 8) are the vertices of isosceles triangle. Calculate its area.
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Answer 1

➡️Vertices of the triangle are A(-5,6), B(3,0), C(9, 8)

✔️Distance between two points=

=√(x^2 - x^1)^2+(y^2 - y^1)^2

AB = √[(3-(-5))^2 + (0-6 )^2] = √64 + 36

= √100

= 10

BC = √[(9-3)^2 + (8-0)^2]

= √36+64

=√100

= 10

AC = √[(9-(-5)^2 + (8-6)^2] = √196 + 4

= √200

= 10√2

AB = BC

Therefore, ΔABC is an isosceles triangle.

(AB)^2 + (BC)^2 = (AC)^2

(10)^2 + (10)^2

= 200 = 10√2 = AC^2

So, the triangle satisfies the pythagoras theorem and hence it is a right angled triangle.

Area of the isosceles right angled triangle

= 1/ 2 × base × height

= 1 /2 *10*10 = 50 sq units.

Hope this helps you.

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Answer 2

u can show them by the help of distance formula.

and for finding area use this formula

of 12 class

1÷2(x1-x2 x2-x3÷y1-2 y2-y3}

hope it will help u

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