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

If $\alpha$ is the angle made by the perpendicular drawn from the origin to the line $3x - 4y + 5 = 0$ with the positive $X$-axis in the positive direction,and $ax + by = 1$ is the equation of a line passing through the point $(1, -1)$ with $\tan \alpha$ as its slope,then $a + ab + b =$

The point on the line $4x - y - 2 = 0$ which is equidistant from the points $(-5, 6)$ and $(3, 2)$ is:

Consider the fourteen lines in the plane given by $y=x+r$ and $y=-x+r$,where $r \in \{0, 1, 2, 3, 4, 5, 6\}$. The number of squares formed by these lines,whose sides are of length $\sqrt{2}$,is:

$A$ square of side $a$ lies above the $x$-axis and has one vertex at the origin. The side passing through the origin makes an angle $\alpha, (0 < \alpha < \frac{\pi}{4})$ with the positive direction of the $x$-axis. The equation of its diagonal not passing through the origin is

The line joining two points $A(2,0)$ and $B(3,1)$ is rotated about $A$ in an anti-clockwise direction through an angle of $15^\circ$. The equation of the line in the new position is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module