Question

The line through $(4,3)$ and $(-6,0)$ intersects the line $5x+y=0$. Find the angles of intersection.

Answer 1

Given that

Point $(x_1,y_1)=(4,3)$

$(x_2,y_2)=(-6,0)$

Line $5x+y=0---------1$

Equation of line

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

$y-3=\frac{0-3}{-6-4}(x-4)$

$y-3=\frac{-3}{-10}(x-4)$

$y-3=\frac{3}{10}(x-4)$

$10y-30=3x-12$

$3x-10y+30=12$

$3x-10y+18=0---------2$

By equation $\left(1\right)$

$5x+y=0$

$y=-5x+0$

Comparing $y=m_{1}x+c$

$m_1=-5$

By equation $\left(2\right)$

$3x-10y+18=0$

$10y=3x+18$

$y=\frac{3}{10}x+\frac{18}{10}$

Comparing $y=m_{2}x+c$

$m_2=\frac{3}{10}$

Then,

Angle, $\tan \theta =\frac{m_2-m_1}{1+m_1{m}_2}$

$=\frac{\frac{3}{10}-(-5)}{1+\frac{3}{10}x(-5)}$

$=\frac{3+50}{\frac{10}{1-6}}$

$=\frac{53\times -5}{10}$

$\tan \theta =-\frac{53}{2}$

Hence this is the answer