If P(0, 7, 10),Q(-1, 6, 6),and R(-4, 9, 6) ar | vedclass

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

Question

(A) Given points are $P(0, 7, 10)$,$Q(-1, 6, 6)$,and $R(-4, 9, 6)$.Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$:$PQ

Vedclass · Accepted Answer

right-angled isosceles triangle

Vedclass · Answer

(A) Given points are $P(0, 7, 10)$,$Q(-1, 6, 6)$,and $R(-4, 9, 6)$.Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$:$PQ = \sqrt{(-1-0)^2 + (6-7)^2 + (6-10)^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$.$QR = \sqrt{(-4 - (-1))^2 + (9-6)^2 + (6-6)^2} = \sqrt{(-3)^2 + 3^2 + 0^2} = \sqrt{9 + 9 + 0} = \sqrt{18} = 3\sqrt{2}$.$PR = \sqrt{(-4-0)^2 + (9-7)^2 + (6-10)^2} = \sqrt{(-4)^2 + 2^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$.Since $PQ = QR = 3\sqrt{2}$,the triangle is isosceles.Also,$PQ^2 + QR^2 = 18 + 18 = 36 = PR^2$. By the converse of the Pythagorean theorem,$	riangle PQR$ is a right-angled triangle.Therefore,$	riangle PQR$ is a right-angled isosceles triangle.

If $P(0, 7, 10)$,$Q(-1, 6, 6)$,and $R(-4, 9, 6)$ are three points in space,then $\triangle PQR$ is:

Similar Questions

If the mid-points of the sides $AB, BC, CA$ of a triangle are $(1, 5, -1), (0, 4, -2), (2, 3, 4)$ respectively,then the length of the median drawn from $C$ to $AB$ is

If $A \equiv (5, 1, p)$,$B \equiv (1, q, p)$,and $C \equiv (1, -2, 3)$ are the vertices of a triangle and $G \equiv (r, -\frac{4}{3}, \frac{1}{3})$ is its centroid,then the values of $p, q, r$ are respectively:

The perimeter of the triangle with vertices at $(1,0,0), (0,1,0)$ and $(0,0,1)$ is

If $G(2, -1, 2)$ is the centroid of tetrahedron $OABC$ where $O=(0, 0, 0)$ and $G_1$ is the centroid of $\triangle ABC$,then $\left|\overline{O G_1}\right|=$

Which of the following statements are true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module