Question

show that the points A(4,6,-5) ,B(0,2,3) and C(-4,-4,-1) form the vertices of an isosceles triangle

Answer 1

Question :-

• Show that the points A(4,6,-5) ,B(0,2,3) and C(-4,-4,-1) form the vertices of an isosceles triangle.

Answer :-

Given :-

• The points A(4,6,-5) ,B(0,2,3) and C(-4,-4,-1)

To Show :-

• Points are vertices of Isosceles Triangle.

Formula used :-

Let us consider two points A and B having coordinates

$\bf \:(x_1,y_1,z_1) \: and \: (x_2,y_2,z_2)$

then distance between A and B is given by

$\bf \:AB = \sqrt{ {(x_2 - x_1)}^{2} + {(y_2 - y_1)}^{2} + {(z_2 - z_1)}^{2} }$

Solution:-

The points A(4,6,-5) ,B(0,2,3) and C(-4,-4,-1)

$\bf \:AB = \sqrt{ {(4 - 0)}^{2} + {(6 - 2)}^{2} + {( - 5 - 3)}^{2} }$

$\bf\implies \:AB = \sqrt{16 + 16 + 64}$

$\bf\implies \:AB = \sqrt{96}$

$\bf \:BC = \sqrt{ {(0 + 4)}^{2} + {(2 + 4)}^{2} + {(3 + 1)}^{2} }$

$\bf\implies \: BC= \sqrt{16 + 36 + 16}$

$\bf\implies \:BC = \sqrt{68}$

$\bf\implies \:CA = \sqrt{ {(4 + 4)}^{2} + {(6 + 4)}^{2} + {( - 5 + 1)}^{2} }$

$\bf\implies \:CA = \sqrt{64 + 100 +16 }$

$\bf\implies \:CA = \sqrt{180}$

Since, not any 2 sides of a triangle are equal.

So, given vertices are not the vertices of Isosceles Triangle.