Name the type of triangle PQR formed by the p | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:$PQ = \sqrt{(-\sqrt{2} - \sqrt{2})^2 + (-\sqrt{2} - \sqrt{2})^2} = \sqrt{(-2

Vedclass · Accepted Answer

Equilateral triangle

Vedclass · Answer

(A) Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:$PQ = \sqrt{(-\sqrt{2} - \sqrt{2})^2 + (-\sqrt{2} - \sqrt{2})^2} = \sqrt{(-2\sqrt{2})^2 + (-2\sqrt{2})^2} = \sqrt{8 + 8} = \sqrt{16} = 4$$PR = \sqrt{(-\sqrt{6} - \sqrt{2})^2 + (\sqrt{6} - \sqrt{2})^2} = \sqrt{(6 + 2 + 2\sqrt{12}) + (6 + 2 - 2\sqrt{12})} = \sqrt{8 + 8} = \sqrt{16} = 4$$RQ = \sqrt{(-\sqrt{6} - (-\sqrt{2}))^2 + (\sqrt{6} - (-\sqrt{2}))^2} = \sqrt{(-\sqrt{6} + \sqrt{2})^2 + (\sqrt{6} + \sqrt{2})^2} = \sqrt{(6 + 2 - 2\sqrt{12}) + (6 + 2 + 2\sqrt{12})} = \sqrt{8 + 8} = \sqrt{16} = 4$Since $PQ = PR = RQ = 4$,all three sides are equal. Therefore,the triangle $PQR$ is an equilateral triangle.

Name the type of triangle $PQR$ formed by the points $P(\sqrt{2}, \sqrt{2})$,$Q(-\sqrt{2}, -\sqrt{2})$,and $R(-\sqrt{6}, \sqrt{6})$.

Similar Questions

Find a point which is equidistant from the points $A(-5, 4)$ and $B(-1, 6)$. How many such points are there?

$P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$ are the given points. If $\overline{PQ}$ is parallel to the $X$-axis,then $PQ = \dots$

In $\Delta ABC$,$A(1, 2)$,$B(3, 4)$,and $C$ lies on the $X$-axis. If the centroid $G$ of $\Delta ABC$ is on the $Y$-axis,then find the coordinates of $C$.

If $A(-2, 1)$,$B(1, 0)$,$C(x, 3)$,and $D(1, y)$ are the vertices of a parallelogram $ABCD$,then find the values of $x$ and $y$.

Find the points of trisection of the line segment joining $(7, 5)$ and $(-2, -1)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module