Question

Let the points p(-1,3),q(6,4) and r(3,0) be the vertical of triangle pqr,then triangle pqr is

Answer 1

Answer:

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Answer 2

As per the data given in the question the vertices are

p(-1,3),q(6,4) and r(3,0)

By calculating the distance between the vertices we can find the sides.

The distance formula is $d = \sqrt{(x_2 - x_1)^2 + (y_2-y_1)^2}$

The distance of pq is $\sqrt{(6-(-1))^{2} +(4-3)^{2} }$

pq=$\sqrt{50}$

The distance of qr is $\sqrt{(3-6)^{2}+(0-4)^{2} }$

$\sqrt{(3)^{2} +(4)^{2} }$=$\sqrt{25}$

The distance of pr is$\sqrt{(-1-3)^{2} +(0-3)^{2} }$

$\sqrt{(-4)^{2} +(-3)^{2} }$=$\sqrt{25} .$

The two sides of the triangle is $\sqrt{25}$ and one side is$\sqrt{50}$.

Therefore ,the triangle is an isosceles triangle.

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