Question

24. A triangle with vertices (4,0),(-1,-1), (3,5), is
(A) isosceles and right angled
B) isosceles but no right ang
(C) right angled but not isosceles​

Answer 1

Answer:-

(a) isosceles and right angled

Explanation:-

Given

Vertices are (4,0), (-1,-1) and (3,5)

To Find

The triangle formed by the given points

Solution

Let,

A(4,0) , B(-1,-1) and C(3,5) are the points or vertices of triangle

Now

From the distance formula

$\boxed{\sqrt{{(x_{2}-x_{1})}^{2} +{(y_{2}-y_{1})}^{2} }}$

Distance between A and B

$AB = \sqrt{{(-1-4)}^{2} +{(-1-0)}^{2} }$

$AB = \sqrt{{(-5)}^{2} +{(-1)}^{2} }$

$AB = \sqrt{25+1 }$

$AB = \sqrt{26 }$ units

Distance between B and C

$BC = \sqrt{{(3+1)}^{2} +{(5+1)}^{2} }$

$BC = \sqrt{{(4)}^{2} +{(6)}^{2} }$

$BC = \sqrt{16+36 }$

$BC = \sqrt{52 }$ units

Distance between A and C

$AC = \sqrt{{(3-4)}^{2} +{(5-0)}^{2} }$

$AC = \sqrt{{(-1)}^{2} +{(5)}^{2} }$

$AC = \sqrt{1 +25 }$

$AC = \sqrt{26 }$ units

From Pythagorus theorem

${BC}^{2} = {AB}^{2} +{AC}^{2}$

${\sqrt{52}}^{2} = {\sqrt {26}}^{2} +{\sqrt{26}}^{2}$

$52 =26 + 26$

$52 = 52$

Since

$\blue{\boxed{\red{AB = AC}}}$

and

$\green{\boxed{\pink{{BC}^{2} = {AB}^{2} +{AC}^{2}}} }$

The triangle formed by the points (4,0),(-1,-1), (3,5) is an isosceles right angled triangle

Answer 2

ANSWER:

The option is (1)

EXPLANATION:

Given vertices of the triangle ABC

A(4,0) B(-1,-1) C(3,5)

Now let us first find the sides of the triangle

AB=

$\sqrt{ {(4 - ( - 1)) }^{2} + {(0 - ( - 1))}^{2} }$

$= \sqrt{26}units$

BC=

$\sqrt{ {( - 1 - 3)}^{2} + {( - 1 - 5)}^{2} }$

$= \sqrt{52} units$

AC=

$\sqrt{ {(4 - 3)}^{2} + {(0 - 5)}^{2} }$

$= \sqrt{26} units$

As two of the sides are equal

it is an isosceles triangle

Now,

If it is right angled triangle it must satisfy the Pythagoras theorem

$\bf{BC}^2={AB}^2+{AC}^2$

${ \sqrt{52} }^{2} = { \sqrt{26} }^{2} + \sqrt{ 26} ^{2}$

$52 = 52$

Therefore

it is a right angled isosceles triangle