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​

Answer 1

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 :

$\longmapsto\tt\boxed{Section\:Formula=x=\dfrac{{m}_{1}{x}_{2}+{m}_{2}{x}_{1}}{{m}_{1}+{m}_{2}}\:\:,\:\:y=\dfrac{{m}_{1}{y}_{2}+{m}_{2}{y}_{1}}{{m}_{1}+{m}_{2}}}[tex]</p><p></p><p> </p><p></p><h2><u>Putting Values :</u></h2><p></p><p>[tex]\longmapsto\tt{0=\dfrac{{k}\times{(-1)}+{1}\times{5}}{k+1}\:\:,\:\:y=\dfrac{{k}\times{(-4)}+{1}\times{-6}}{k+1}}$

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

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 .

Answer 2

Let the line segment A(5, -6) and B(-1, -4) is divided at point P(0, y) by y-axis in ratio m:n

:. x = $\frac{mx2+nx1}{m+n}$ and y = $\frac{my2+ny1}{m+n}$

Here, (x, y) = (0, y); (x1, y1) = (5, -6) and (x2, y2) = (-1, -4)

So , 0 = $\frac{m(-1)+n(5)}{m+n}$

=> 0 = -m + 5n

=> m= 5n

=> $\frac{m}{n}$ = $\frac{5}{1}$

=> m:n = 5:1

Hence, the ratio is 5:1 and the division is internal.Now,

y = $\frac{my2+ny1}{m+n}$

=> y = $\frac{5(-4)+1(-6)}{5+1}$

=> y = $\frac{-20-6}{6}$

=> y = $\frac{-26}{6}$

=> y = $\frac{-13}{3}$

Hence, the coordinates of the point of division is (0, -13/3).

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