The triangle with vertices at (-2, 2),(2, -2) | vedclass

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

Question

(C) Let $A(-2, 2)$,$B(2, -2)$,and $C(1, 1)$ be the vertices of $	riangle ABC$.Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:$

Vedclass · Accepted Answer

isosceles triangle

Vedclass · Answer

(C) Let $A(-2, 2)$,$B(2, -2)$,and $C(1, 1)$ be the vertices of $	riangle ABC$.Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:$AB = \sqrt{(2 - (-2))^2 + (-2 - 2)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$$BC = \sqrt{(1 - 2)^2 + (1 - (-2))^2} = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}$$AC = \sqrt{(1 - (-2))^2 + (1 - 2)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$Since $BC = AC = \sqrt{10}$,two sides of the triangle are equal.Therefore,$	riangle ABC$ is an isosceles triangle.

The triangle with vertices at $(-2, 2)$,$(2, -2)$,and $(1, 1)$ is:

Similar Questions

Let $P$ be an interior point of a convex quadrilateral $ABCD$ and $K, L, M, N$ be the mid-points of $AB, BC, CD, DA$ respectively. If $\text{Area}(PKAN) = 25$,$\text{Area}(PLBK) = 36$,and $\text{Area}(PMDN) = 41$,then $\text{Area}(PLCM)$ is

The circumcentre of the triangle formed by the lines $x = -2$,$y = -2$ and $x + y + 2 = 0$ is

The points $(0, 0)$,$(a, 0)$,and $\left( \frac{a}{2}, \frac{a\sqrt{3}}{2} \right)$ are vertices of:

If the vertices of a triangle have integral coordinates,then the triangle is

In an isosceles right-angled triangle,if the equation of the hypotenuse is $3x + 4y = 4$ and its opposite vertex is $(2, 2)$,then the slopes of the remaining two sides are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module