Question

Find the ratio in which the line joining $(-2,5)$ and $(-5,-6)$ is divided by the line $y=-3$. Hence find the point of intersection.

Answer 1

$\to$ Let $x_1,y_1=(-2,5)$

$x_2,y_2=(-5,-6)$

And here $y=(-3)$

$\therefore y=\frac{m_1{y}_2+m_2{y}_1}{m-\left[1\right]+m_2}$

$\therefore (-3)=\frac{m_1(-6)+m_2\left(5\right)}{m_1+m_2}\to \left(1\right)$

$\therefore -3m_1-3m_2=-6m_1+5m_2$

$\therefore -3m_1+6m_1=5m_2+3m_2$

$\therefore 3m_1=8m_2$

$\therefore \frac{m_1}{m_2}=\frac{8}{3}$

$\to$ The ratio will be $m_1:m_2=8:3$

$\to$ To find equation of line passing through the above co-ordinates we use intercept forms:

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

$\therefore y-5=\frac{-6-5}{-5(-2)}(x-(-2\left)\right)$

$\therefore y-5=\frac{-11}{-3}(x+2)$

$\therefore y-5=\frac{11}{3}(x+2)$

$\therefore 3y-15=11x+22$

$\therefore 3y=11x+37$

$\therefore y=\frac{11}{3}x+\frac{37}{3}\Rightarrow$ equation of another line

$\to$ To find intersection of both the line

$11x+37=3y$

$\therefore 11x-3y+37=0\to \left(1\right)$

$y+3=0\to \left(2\right)$

$\therefore$ From eq $\left(2\right)$ we get

$y=(-3)$

$\to$ Putting value of $y$ in eq $\left(1\right)$

$11x-3(-3)+37=0$

$\therefore 11x+9+37=0$

$\therefore x=\left(\frac{-46}{11}\right)$

Therefore point of intersection

$(\frac{-46}{11},-3)$