Question

check whether (5,-2),(6,4)and(7-2) are the vertices of an isoceles triangle.​

Answer 1

Answer:

Yes they form an isoceles triangle.

Explanation:

Let the given points be -

• A(5, -2)
• B(6, 4)
• C(7, -2)

Since, it forms an isoceles triangle distance between any two vertices must be equal.

Distance of AB is given by,

$\longrightarrow \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Where,

• $x_1 \: {\rm and } \: x_2 \rm \: is \: 5 \: and 6$
• And, $y_1 \: {\rm and } \: y_2 \rm \: is \: -2 \: and 4$

So,

$\to \: \sqrt{ {(6 - 5)}^{2} + { \big(4 - ( - 2) \big)}^{2} }$

$\to \: \sqrt{ {(1)}^{2} + ( 4+ 2 {)}^{2} }$

$\to \: \sqrt{1 + {6}^{2} }$

$\to \: \sqrt{1 + 36}$

$\to \rm \sqrt{37} \: units$

So on, the distance of BC :-

$\to \: \sqrt{( x_2 - x_1 {)}^{2} + {(y_2 - y_1)}^{2} }$

$\to \: \sqrt{ {(7 - 6)}^{2} + { \big(4 - ( - 2) \big)}^{2} }$

$\to \: \sqrt{ {(1)}^{2} + {(4 + 2)}^{2} }$

$\to \: \sqrt{1 + {(6)}^{2} }$

$\to \: \sqrt{1 + 36}$

$\to \: \rm\sqrt{37} \: units.$

And also, calculating distance of AC :-

$\to \: \sqrt{( x_2 - x_1 {)}^{2} + {(y_2 - y_1)}^{2} }$

$\to \: \sqrt{(7 - 5 {)}^{2} + { \big( - 2 - ( - 2) \big)}^{2} }$

$\to \: \sqrt{ {(2)}^{2} + ( - 2 + 2)}$

$\to \: \sqrt{4 + 0}$

$\to \rm \: 2 \: units$

So here, we are getting two units equal AB and BC as √37 units, which proves the property of an isoceles traingle i.e., An isoceles traingle have two sides equal.

Hence, given points (5, -2), (6, 4) and (7, -2) are the vertices of a isoceles traingle.

_____________________

Formula used :

$\to \: \sqrt{( x_2 - x_1 {)}^{2} + {(y_2 - y_1)}^{2} }$

• $(x_1, y_1)$ denotes coordinates of the first point.

• $(x_2, y_2)$ denotes coordinates of the second point.