Question

A vertex of an equilateral triangle is (2, 3) and the equation of the opposite side is x + y = 2. Find the equation of the other sides of the triangle.

• A.
• B.
• C.
• D.

Answer 1

Answer: (A), (D)

The correct options are
A

D

We know that lines of equlateral triangle are inclined at ${60}^o$.

Slope of line BC=-1
Compare given line is x +y =2 with y =mx +c
Slope of line BC =-1
Let the slope of line $AB=m_1$ and slope of line $AC=m_2$
Angle between line AB and AC
$tan\theta =|\frac{m_1-m_2}{1+m_1{m}_2}|$

$tan{60}^o=|\frac{m_1-(-1)}{1+m_1(-1)}|$

$\sqrt{3}=|\frac{m_1+1}{1-m_1}|$
$\sqrt{3}=\frac{m_1+1}{1-m_1}or-\sqrt{3}=\frac{m_1+1}{1-m_1}$

$\sqrt{3}-\sqrt{3}{m}_1=m_1+1or-\sqrt{3}+\sqrt{3}{m}_1=m_1+1$
$\sqrt{3}-1=m_1(1+\sqrt{3})or-(\sqrt{3}-1)m_1=\sqrt{3}+1$
$m_1=\frac{\sqrt{3}-1}{\sqrt{3}+1}or{m}_1=\frac{\sqrt{3}+1}{\sqrt{3}-1}$
$m_1=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}or{m}_1=\frac{\sqrt{3}+1}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
$\frac{3+1-2\sqrt{3}}{3-1}\frac{3+1+2\sqrt{3}}{3-1}$
$m_1=2-\sqrt{3}$
Slope of on line is $m_1=2-\sqrt{3}$ and slope of other line is $m_2+\sqrt{3}$ both the lines passes through (2,3)

Equation of line 1
$y-y_1=m(x-x_1)$
$y-3=(2-\sqrt{3})x-4+m_1=2\sqrt{3}$
$(2-\sqrt{3})x-y+m_1=2\sqrt{3}-1=0$

Equation of line 2
$Y-3=2+\sqrt{3}(x-2)$
$Y-3=(2+\sqrt{3})x-4-2\sqrt{3}$
$(2+2\sqrt{3})x-y-2\sqrt{3}-1=0$
So, options A and D are correct