Find angles between the lines $\sqrt3x+y=1$ and $x+\sqrt3y=1.$

Question

If a point $ P(x, y) $ satisfies the condition that its distance from the point $ (3, -2) $ is equal to its distance from the line $ y = 2x + 1 $, then the locus of point $ P $ is:

Answer 1

The given lines are $\sqrt3x+y=1$ and $x+\sqrt3y=1.$

$y = -\sqrt3x + 1 … (1)$ and $y = \frac{-1}{\sqrt3x} +\frac{ 1}{\sqrt3} …. (2)$

The slope of line (1) is $m_1 = -\sqrt3$ , while the slope of line (2) is $m_2 =\frac{ -1}{\sqrt3}$
The acute angle i.e., $θ$ between the two lines is given by

$Tanθ=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$

$Tanθ=\left|\frac{-\sqrt3+\frac{1}{\sqrt3}}{1+(-\sqrt3)(\frac{-1}{\sqrt3})}\right|$

$Tanθ=\left|\frac{\frac{-3+1}{\sqrt3}}{1+1}\right|$

$=\left|\frac{-2}{2\times\sqrt3}\right|$

$Tanθ=\frac{1}{\sqrt3}$

$θ=30º$
Thus, the angle between the given lines is either $ 30° $or $180° - 30° = 150°$.

Find angles between the lines \(\sqrt3x+y=1\) and \(x+\sqrt3y=1.\)

Solution and Explanation

Top Questions on Distance of a Point From a Line