Question

A (6,–4), B (–2, –4), C (2,10) Show that the points taken in order form an isosceles triangle.

Answer 1

Hi ,

Given points are A(6,-4),B(-2,-4)

and C(2,10)

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The distance between points

P(x1, y1) and Q(x2,y2) is

PQ = √(x2 - x1)² + ( y2 - y1)²

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i )Distance between A(6,-4) and B(-2,-4)

AB = √[-2-6 ]² + [ -4 - ( -4 )]²

= √ (-8)²+ 0

AB = 8

ii )Distance between B(-2,-4) and C(2,10)

BC= √[ 2 - (-2 )]² + [ 10 - (-4)]²

= √ 4² + 14²

= √ 16 + 196

BC = √212

iii ) distance between C(2,10) and A(6,-4)

CA = √(6-2)² + ( -4 -10 )²

= √ 4² + 14²

= √ 16 + 196

= √212

BC = CA = √212

Therefore ,

In ∆ABC , BC = CA

ABC is an isosceles triangle.

I hope this helps you.

: )

Answer 2

Hello,

Solution:

Distance Formula between two points

A(x1, y1) B(x2,y2)

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

A(6,-4) B(-2,-4)

$AB = \sqrt{( { - 2 - 6)}^{2} + ( { - 4 + 4)}^{2} } \\ \\ = \sqrt{64} \\ \\ = 8$

B(-2,-4) C(2,10)

$BC = \sqrt{ {( - 2 - 2)}^{2} + ( { - 4 - 10)}^{2} } \\ \\ = \sqrt{16 + 196} \\ \\ = \sqrt{212} \\ \\ = 2 \sqrt{53}$

C(2,10) A(6,-4)

$AC = \sqrt{ {(2 - 6)}^{2} + ( {10 + 4)}^{2} } \\ \\ = \sqrt{16 + 196} \\ \\ = \sqrt{212} \\ \\ = 2 \sqrt{53}$

Since in Isosceles Triangle two sides are equal ,so here distances BC = CA

Hope it helps you.