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Mathematics Question on Distance of a Point From a Line

Two lines passing through the point (2, 3) intersects each other at an angle of 60º.60º . If slope of one line is 2, find equation of the other line.

Answer

It is given that the slope of the first line, m1=2.m_1 = 2.
Let the slope of the other line be m2m_2.
The angle between the two lines is 60°.60°.

∴tan60º=∣m1−m21+m1m2∣∴tan60º=\left|\frac{m_1-m_2}{1+m_1m_2}\right|

⇒3=∣2−m21+2m2∣⇒\sqrt3=\left|\frac{2-m_2}{1+2m_2}\right|

⇒3=±(2−m21+2m2)⇒\sqrt3=±\left(\frac{2-m_2}{1+2m_2}\right)

⇒3=(2−m21+2m2)or3=−(2−m21+2m2)⇒\sqrt3=\left(\frac{2-m_2}{1+2m_2}\right) or \sqrt3=-\left(\frac{2-m_2}{1+2m_2}\right)

⇒3(1+2m2)=2−m2  or3(1+2m2)=−(2−m2)⇒\sqrt3(1+2m_2)=2-m_2 \space or \sqrt3(1+2m_2)=-(2-m_2)
⇒3+23m2+m2=2  or3+23m2−m2=−2⇒\sqrt3+2\sqrt3m_2+m_2=2 \space or \sqrt3+2\sqrt3m_2-m_2=-2

⇒m2=2−3(23+1)  or  m2=−(2+3)(23−1)⇒m_2=\frac{2-\sqrt3}{\left(2\sqrt3+1\right)}\space or\space m_2=-\frac{\left(2+\sqrt3\right)}{\left(2\sqrt3-1\right)}

Case I : m2=2−3(23+1)m_2=\frac{2-\sqrt3}{\left(2\sqrt3+1\right)}
The equation of the line passing through point (2, 3) and having a slope of(2−3)(23+1)\frac{\left(2-\sqrt3\right)}{\left(2\sqrt3+1\right)} is

(y−3)=2−323+1(x−2)(y-3)=\frac{2-\sqrt3}{2\sqrt3+1}(x-2)

(23+1)(y−3)=(2−3)x−2(2−3)(2\sqrt3+1)(y-3)=(2-\sqrt3)x-2(2-\sqrt3)

(3−2)x+(23+1)y=−4+23+63+3(\sqrt3-2)x+(2\sqrt3+1)y=-4+2\sqrt3+6\sqrt3+3
(3−2)x+(23+1)y=−1+83(\sqrt3-2)x+(2\sqrt3+1)y=-1+8\sqrt3
In this case, the equation of the other line is (3−2)x+(23+1)y=−1+83(\sqrt3-2)x+(2\sqrt3+1)y=-1+8\sqrt3

Case II : m2=−(2+3)(23−1)m_2=\frac{-\left(2+\sqrt3\right)}{\left(2\sqrt3-1\right)}
The equation of the line passing through point (2, 3) and having a slope of−(2+3)(23−1) \frac{-\left(2+\sqrt3\right)}{\left(2\sqrt3-1\right)} is

(y−3)=−(2+3)(23−1)(x−2)(y-3)=\frac{-\left(2+\sqrt3\right)}{\left(2\sqrt3-1\right)}\left(x-2\right)

(23−1)y−3(23−1)=−(2+3)x+2(2+3)(2\sqrt3-1)y-3(2\sqrt3-1)=-(2+\sqrt3)x+2(2+\sqrt3)
(23−1)y+(2+3)x=4+23+63−3(2\sqrt3-1)y+(2+\sqrt3)x=4+2\sqrt3+6\sqrt3-3
(2+3)x+(23−1)y=1+83(2+\sqrt3)x+(2\sqrt3-1)y=1+8\sqrt3

In this case, the equation of the other line is(2+3)x+(23−1)y=1+83 (2+\sqrt3)x+(2\sqrt3-1)y=1+8\sqrt3

Thus, the required equation of the other line is (\sqrt3-2)x+(2\sqrt3+1)y$$=-1+8\sqrt3 or (2+3)x+(23−1)y=1+83 (2+\sqrt3)x+(2\sqrt3-1)y=1+8\sqrt3