Question

name the type of triangle PQR formed by the point p(√2,√2) ,q(-√2,-√2),R(-√6,√6)

Answer 1

first we find length of all sides of triangle.
e.g.
PQ=√{(2√2)^2+(2√2)^2}=4
QR=√{(√2+√6)^2+(√2-√6)^2}=4
RP=√{(√2+√6)^2+(√2-√6)^2}=4
here we see
PQ=QR=RP
hence ∆PQR is an equilateral triangle

Answer 2

Answer:

Equilateral triangle

Step-by-step explanation:

P=(√2,√2)

Q=(-√2,-√2)

R=(-√6,√6)

To find PQ use distance formula :

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

$(x_1,y_1)=(\sqrt{2},\sqrt{2})$

$(x_2,y_2)=(-\sqrt{2},-\sqrt{2})$

Substitute the values in the formula :

$PQ=\sqrt{(-\sqrt{2}-\sqrt{2})^2+(-\sqrt{2}-\sqrt{2})^2}$

$PQ=\sqrt{(-2\sqrt{2})^2+(-2\sqrt{2})^2}$

$PQ=\sqrt{8+8}$

$PQ=\sqrt{16}$

$PQ=4$

To Find QR

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

$(x_1,y_1)=(-\sqrt{2},-\sqrt{2})$

$(x_2,y_2)=(-\sqrt{6},\sqrt{6})$

Substitute the values in the formula :

$QR=\sqrt{(-\sqrt{6}+\sqrt{2})^2+(\sqrt{6}+\sqrt{2})^2}$

$QR=\sqrt{16}$

$QR=4$

To Find PR

$(x_1,y_1)=(\sqrt{2},\sqrt{2})$

$(x_2,y_2)=(-\sqrt{6},\sqrt{6})$

Substitute the values in the formula :

$PR=\sqrt{(-\sqrt{6}-\sqrt{2})^2+(\sqrt{6}-\sqrt{2})^2}$

$PR=\sqrt{16}$

$PR=4$

Thus the sides of the triangle is PQ=QR=PR = 4 cm

Hence the given triangle is an equilateral triangle