Question

prove that the points A(-3,0), B(1,-3 ) and (4,1) are the vertices of an isosceles right angle triangle. find the area of the triangle.​

Answer 1

Solution :-

A( - 3, 0 ) B( 1, - 3 ) C( 4, 1 )

Let us find measure if each side of the triangle

i) A( - 3, 0 ) B( 1, - 3)

$\sf Distance ( d ) = \sqrt{ {( x_2 - x_1) }^{2} + {(y_2 - y_1) }^{2} }$

$\implies \sf AB= \sqrt{ { \{1 - ( - 3) \} }^{2} + {( - 3 - 0) }^{2} }$

$\implies \sf AB= \sqrt{ { (1 + 3) }^{2} + {( - 3 ) }^{2} }$

$\implies \sf AB= \sqrt{ { 4 }^{2} + 9 }$

$\implies \sf AB= \sqrt{ 16+ 9 }$

$\implies \sf AB= \sqrt{25} = 5 \ units$

ii ) B( 1, - 3 ) C( 4, 1 )

$\sf Distance ( d ) = \sqrt{ {( x_2 - x_1) }^{2} + {(y_2 - y_1) }^{2} }$

$\implies \sf BC= \sqrt{ { (4 - 1) }^{2} + { \{1 - ( - 3) \} }^{2} }$

$\implies \sf BC= \sqrt{ { 3 }^{2} + { (1 + 3) }^{2} }$

$\implies \sf BC= \sqrt{ 9 + 4^{2} }$

$\implies \sf BC= \sqrt{ 9 + 16 }$

$\implies \sf BC= \sqrt{ 25 } = 5 \ units$

iii ) A( - 3, 0 ) C( 4, 1 )

$\sf Distance ( d ) = \sqrt{ {( x_2 - x_1) }^{2} + {(y_2 - y_1) }^{2} }$

$\implies \sf AC = \sqrt{ { \{4 - ( - 3) \}}^{2} + { (1 - 0) }^{2} }$

$\implies \sf AC = \sqrt{ {7}^{2} + 1 }$

$\implies \sf AC = \sqrt{ 49 + 1 } = \sqrt{50} \ units$

AB = BC = 5 units

We know that

A triangle in which any 2 sides are equal in length is known as an Isosceles triangle.

Therefore, ΔABC is an Isosceles triangle

Hence proved.

Finding the area of the triangle

If we observe

5² + 5² = 25 + 25 = 50 = ( √50 )²

i.e, 5² + 5² = ( √50 )²

i.e AB² + BC² = AC²

i.e, Sum of sides of lengths of 2 sides is equal to square of length of 3rd side.

So, By Pythagoras Theorem

ΔABC is a Right angled triangle

In a Right angled isosceles triangle Base and Height are equal

Base ( AB ) = 5 units

Height ( BC ) = 5 units

ar( ΔABC ) = ( 1/2 ) * Base * Height

→ ar( ΔABC ) = ( 1/2 ) * AB * BC

→ ar( ΔABC ) = ( 1/2 ) * 5 * 5

→ ar( ΔABC ) = ( 1/2 ) * 25

→ ar( ΔABC ) = 25/2 = 12.5 sq.units

Hence, the area of the triangle is 12.5 sq.units.