Question

find the is of the straight line passing
through the point
(1, 2) & making
angle of 60" with the line 13x +y +210​

Answer 1

✪ᴀɴsᴡᴇʀ✪

• Equations of lines are

$\:\:\:\:\:\:\: \boxed{\bf{\red{ y - 2 = \left[\frac{-13+ \sqrt{3}}{13\sqrt{3} + 1}\right](x-1)}}}$

$\:\:\:\:\:\:\: \boxed{\bf{\green{y - 2 = \left[\frac{13 + \sqrt{3}}{13\sqrt{3} - 1}\right](x-1)}}}$

☆ɢɪᴠᴇɴ☆

• A line pass through (1,2) and makes an angle of $60^o$ with the line $13x+y+21 = 0$

★ᴛᴏ ғɪɴᴅ★

• Equation of the line.

⍟ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴ⍟

Let $m_1, m_2$ represents slope of the given line $13x+y+21 = 0$ and the required line respectively.

Now

$\:\:\:\: m_1 = - 13$

Angle $\theta$ between two lines is given by

$\to \left| \frac{m_1-m_2} {1+m_1m_2} \right| = \tan(\theta)$

$\to \left| \frac{-13-m_2} {1-13m_2} \right| = \tan(60^o)$

$\to \left| \frac{m_2+13} {1-13m_2} \right| = \sqrt{3}$

$\to \frac{m_2+13} {1-13m_2} = +\sqrt{3}$

$or\:\frac{m_2+13} {1-13m_2} = - \sqrt{3}$

$\to m_2+13 = \sqrt{3}(1-13m_2)$

$or\: m_2+13= - \sqrt{3}(1-13m_2)$

$\to 13\sqrt{3}m_2 + m_2 = -13+ \sqrt{3}$

$or\: 13\sqrt{3}m_2 - m_2 = 13 + \sqrt{3}$

$\to (13\sqrt{3} + 1)m_2 = -13+ \sqrt{3}$

$or\: (13\sqrt{3} - 1)m_2 = 13 + \sqrt{3}$

$\to m_2 = \frac{-13+ \sqrt{3}}{13\sqrt{3} + 1}$

$or\: m_2 = \frac{13 + \sqrt{3}}{13\sqrt{3} - 1}$

Using Point Slope Form of line, we get

$\:\:\:\: y - y_1 = m_2(x - x_1)$

$\:\:\:\: y - 2 = m_2(x - 1)$

Substituting values of $m_2$, the two equations of lines are

$\to y - 2 = \left[\frac{-13+ \sqrt{3}}{13\sqrt{3} + 1}\right](x-1)$

$or\: y - 2 = \left[\frac{13 + \sqrt{3}}{13\sqrt{3} - 1}\right](x-1)$