Question

find the equation of the line passing through the point (2,3) and making an angle of 45° with the line 3x+ y -5 =0

Answer 1

Question :-

→ Given above ↑

Answer :-

$\mapsto \boxed{\sf \: y - 6x + 9 = 0} \:$

To Find :-

→ The equation of line passing through given points and making angle with the given line .

Solution :-

We have ,

→ Equation of line

$\to \sf \: 3x + y - 5 = 0 \\ \\ \sf differentiate \: with \: respect \: to \: x \\ \\ \to \sf \: 3 + \frac{dy}{dx} = 0 \\ \\ \to \boxed{ \sf \: slope \: (m_1 ) \: \frac{dy}{dx} = - 3}$

Let ,the slope of required line is "m"

according to the question required line makes 45° with another line so ,

$\to \tan 45 = \sf \bigg| \frac{m - {3} }{3} \bigg| \: \\ \\ \to \sf 3 =| m \: - 3 | \: \\ \\ \to \boxed{\sf | m |= 6}$

We know that

Equation of line passing through the point $\sf (x_1 ,y_1)$ with slope m is -

$\to \boxed{ \sf y - y_1 = m(x - x_1)}$

Hence , required equation of line which passing through (2,3) with slope "6" is -

$\mapsto \sf \: y - 3 = 6(x - 2) \\ \\ \mapsto \sf \: y - 3 = 6x - 12 \\ \\ \mapsto \sf y - 6x - 3 + 12 = 0 \\ \\ \mapsto \boxed{\sf \: y - 6x + 9 = 0}$