One vertex of an equilateral triangle is (2, | vedclass

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

Question

(C) Let the slope of the required lines be $m$. The given line is $x + y = 2$,which has a slope of $m_1 = -1$.Since the triangle is equilateral,the an

Vedclass · Accepted Answer

$y - 3 = (2 \pm \sqrt{3})(x - 2)$

Vedclass · Answer

(C) Let the slope of the required lines be $m$. The given line is $x + y = 2$,which has a slope of $m_1 = -1$.Since the triangle is equilateral,the angle between the sides is $60^\circ$.Using the formula $	an 	heta = |\frac{m - m_1}{1 + m \cdot m_1}|$,we have:$	an 60^\circ = |\frac{m - (-1)}{1 + m(-1)}|$$\sqrt{3} = |\frac{m + 1}{1 - m}|$This gives two cases:$1) \sqrt{3} = \frac{m + 1}{1 - m}$ $\Rightarrow \sqrt{3} - \sqrt{3}m = m + 1$ $\Rightarrow m(1 + \sqrt{3}) = \sqrt{3} - 1$ $\Rightarrow m = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = 2 - \sqrt{3}$$2) -\sqrt{3} = \frac{m + 1}{1 - m}$ $\Rightarrow -\sqrt{3} + \sqrt{3}m = m + 1$ $\Rightarrow m(\sqrt{3} - 1) = \sqrt{3} + 1$ $\Rightarrow m = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = 2 + \sqrt{3}$Using the point-slope form $y - y_1 = m(x - x_1)$ with point $(2, 3)$:$y - 3 = (2 \pm \sqrt{3})(x - 2)$.

One vertex of an equilateral triangle is $(2, 3)$ and the line of its opposite side is $x + y = 2$. Find the equations of the other two sides.

Similar Questions

$A(-2, 3)$ is a point on the line $4x + 3y - 1 = 0$. If the points on the line that are $10$ units away from the point $A$ are $(x_1, y_1)$ and $(x_2, y_2)$,then $(x_1 + y_1)^2 + (x_2 + y_2)^2 =$

If $l, m, n$ are in arithmetic progression,then the straight line $lx + my + n = 0$ will always pass through the point:

Let the distance between two parallel lines be $5$ units and a point $P$ lie between the lines at a unit distance from one of them. An equilateral triangle $PQR$ is formed such that $Q$ lies on one of the parallel lines,while $R$ lies on the other. Then $(QR)^2$ is equal to . . . . . . .

The distance of the point $(1, 2)$ from the line $x + y = 0$ measured parallel to the line $3x - y = 2$ is

$L_1 \equiv 2x+y-3=0$ and $L_2 \equiv ax+by+c=0$ are two equal sides of an isosceles triangle. If $L_3 \equiv x+2y+1=0$ is the third side of this triangle and $(5,1)$ is a point on $L_2=0$,then $\frac{b^2}{|ac|}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module