Question

solve the attachment​

Answer 1

Correct Question :

Find the ratio in which the y-axis divides the line segment joining the points (5,-6) and (-1,-4). Also find the point of intersection.

Given :

Two points (5,-6) and (-1,-4)

To Find :

The ratio in which the y-axis divides the line segment joining the points (5,-6) and (-1,-4).

Solution :

$\longmapsto\tt{Let\:A=(5,-6)}$

$\longmapsto\tt{Let\:B=(-1,-4)}$

Using Formula :

$\longmapsto\tt\boxed{Section\:Formula=x=\dfrac{{m}_{1}{x}_{2}+{m}_{2}{x}_{1}}{{m}_{1}+{m}_{2}}\:\:,\:\:y=\dfrac{{m}_{1}{y}_{2}+{m}_{2}{y}_{1}}{{m}_{1}+{m}_{2}}}$

Putting Values :

$\longmapsto\tt{0=\dfrac{{k}\times{(-1)}+{1}\times{5}}{k+1}\:\:,\:\:y=\dfrac{{k}\times{(-4)}+{1}\times{-6}}{k+1}}$

$\longmapsto\tt{0=\dfrac{-k+5}{k+1}\:,\:y=\dfrac{-4k-6}{k+1}}$

Now ,

By Cross Multiply :

$\longmapsto\tt{0(k+1)=-k+5}$

$\longmapsto\tt{0=-k+5}$

$\longmapsto\tt\bf{k=5}$

Ratio is 5:1 .

Putting k = 5 in (y = -4k-6 / k+1) :

$\longmapsto\tt{y=\dfrac{-4(5)-6}{5+1}}$

$\longmapsto\tt{y=\dfrac{-20-6}{6}}$

$\longmapsto\tt{y=\cancel\dfrac{-26}{6}}$

$\longmapsto\tt\bf{y=\dfrac{-13}{3}}$

So , The Point of Intersection is 0 , -13/3 .