If the points (1, 1),(-1, -1) and (-sqrt{3}, | vedclass

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

Question

(C) Let the vertices be $A(1, 1)$,$B(-1, -1)$,and $C(-\sqrt{3}, k)$.For an equilateral triangle,the distance between any two points must be equal.Firs

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(C) Let the vertices be $A(1, 1)$,$B(-1, -1)$,and $C(-\sqrt{3}, k)$.For an equilateral triangle,the distance between any two points must be equal.First,calculate the length of side $AB$:$AB^2 = (-1 - 1)^2 + (-1 - 1)^2 = (-2)^2 + (-2)^2 = 4 + 4 = 8$.Now,calculate the distance $AC^2$:$AC^2 = (-\sqrt{3} - 1)^2 + (k - 1)^2 = (\sqrt{3} + 1)^2 + (k - 1)^2 = 3 + 1 + 2\sqrt{3} + (k - 1)^2 = 4 + 2\sqrt{3} + (k - 1)^2$.Since $AB^2 = AC^2$,we have:$8 = 4 + 2\sqrt{3} + (k - 1)^2$$(k - 1)^2 = 4 - 2\sqrt{3} = (\sqrt{3} - 1)^2$.Taking the square root,$k - 1 = \pm(\sqrt{3} - 1)$.If $k - 1 = \sqrt{3} - 1$,then $k = \sqrt{3}$.If $k - 1 = -(\sqrt{3} - 1) = 1 - \sqrt{3}$,then $k = 2 - \sqrt{3}$.Comparing with the given options,the correct value is $k = \sqrt{3}$.

If the points $(1, 1)$,$(-1, -1)$ and $(-\sqrt{3}, k)$ are vertices of an equilateral triangle,then the value of $k$ will be:

Similar Questions

The centroid of the triangle formed by the lines $x+y-1=0$,$x-y-1=0$,and $x-3y+3=0$ is

The sides of a triangle are $3x + 4y$,$4x + 3y$,and $5x + 5y$ units,where $x, y > 0$. The triangle is:

In a $\triangle ABC$,$AB = AC = 37$. Let $D$ be a point on $BC$ such that $BD = 7$ and $AD = 33$. The length of $CD$ is:

The in-centre of the triangle formed by the lines $x=0, y=0$ and $3x+4y=12$ is

The diagonals $AC$ and $BD$ of a rhombus $ABCD$ intersect at the point $O(3,4)$. If $BD=2 \sqrt{2}$, $A=(1,2)$, $B=(\alpha, \beta)$, $D=(\gamma, \delta)$ and $\alpha < \delta < \gamma < \beta$, then the value of $\beta+\gamma-\delta$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module