Question

O Examire whether point (2,3), (-4,-1) & (3,-5) are thos
vectices of an isoscelos tiangle
​

Answer 1

Answer:

yes, the points are the vertices of an isosceles triangle.

Step-by-step explanation:

distance formula : √(x2-x1)^2 +(y2-y1)^2

Isosceles triangle : the triangle in which only two sides are equal.

let's take ABC as an isosceles triangle.

A(2,3) ; B(-4,-1) ; C(3,-5)

AB = √(-4-2)^2 + (-1-3)^2

= √(-6)^2 + (-4)^2

= √36 + 16

= √52 = 2√13

BC = √{3-(-4)}^2 + {-5-(-1)}^2

= √(3+4)^2 + (-5+1)^2

= √(7)^2 + (4)^2

= √49 + 16

= √65

AC = √(3-2)^2 + (-5-3)^2

= √(1)^2 + (-8)^2

= √1 + 64

= √65

from the above solution it is clear that the side BC and AC are equal in distance.

so the points are the vertices of an isosceles triangle.

Answer 2

$\large\underline{\bold{Given \:Question - }}$

• Examine whether the points (2,3), (-4,-1) & (3,-5) are the vectices of an isosceles triangle or not.

$\large\underline{\bold{Solution-}}$

• Let us consider a triangle ABC having coordinates A(2,3), B(- 4, - 1) and C(3, - 5)

Concept Used :-

Distance Formula :-

Let us consider two points A and B, then distance D between A and B is given by

$\rm D = \sqrt{ {(x_{2} - x_{1}) }^{2} + {(y_{2} - y_{1})}^{2} }$

$\sf \:where \: coordinates \: are \: A(x_1,y_1) \: and \: B(x_2,y_2)$

Now, we have given here three coordinates A, B and C, we first find the distance AB, BC and CA by using distance formula, if any 2 distances are same, the triangle become isosceles triangle.

Let's now solve the problem!!

Step :- 1

• Distance between A (2, 3) and B (- 4, - 1)

Here,

• • x₁ = 2

• • x₂ = - 4

• • y₁ = 3

• • y₂ = - 2

• So, distance between these points is,

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

$\sf \: AB = \sqrt{ {( - 6)}^{2} + {( - 4)}^{2} }$

$\sf \:AB = \sqrt{36 + 16}$

$\therefore \: \: \boxed{ \bf{AB = \sqrt{52} = 2 \sqrt{13 \:} \: units}}$

Step :- 2

• Distance between the points B(- 4, - 1) and C (3, - 5).

Here,

• • x₁ = - 4

• • x₂ = 3

• • y₁ = - 1

• • y₂ = - 5

• So, distance between these points is,

$\bf \: BC = \sqrt{ {(3 + 4)}^{2} + ( - 5 + 1)^{2} }$

$\sf \: BC = \sqrt{ {(7)}^{2} + {( - 4)}^{2} }$

$\sf \:BC = \sqrt{49 + 16}$

$\therefore \: \: \boxed{ \bf{BC = \sqrt{65} \: units }}$

Step :- 3

• Distance between the points C(3, - 5) and A(2, 3)

Here,

• • x₁ = 3

• • x₂ = 2

• • y₁ = - 5

• • y₂ = 3

• So, distance between these points is,

$\bf \:CA = \sqrt{ {(2 - 3)}^{2} + {(3 + 5)}^{2} }$

$\sf \: CA = \sqrt{ {( - 1)}^{2} + {(8)}^{2} }$

$\sf \: CA = \sqrt{1 + 64}$

$\therefore \: \: \boxed{ \bf{CA = \sqrt{65} }} - - - - (3)$

$\therefore \: \: \boxed{ \bf{BC \: = \: CA \: = \sqrt{65} \: units }}$

$\: \: \boxed{ \bf{Hence\: triangle \: is \: isosceles \: as \: BC =CA}}$