Question

Given a line segment $AB$ joining the points $A(–4,6)$ and $B(8,–3)$. Find:
(i) the ratio in which $AB$ is divided by the $y$-axis.
(ii) find the coordinates of the point of intersection.
(iii) the length of $AB$.

Answer 1

(i) Let $m:n$ be the ratio in which the line segment joining $A(-4,6)$ and $B(8,-3)$ is divided by the $Y$ axis.

Since the line meets $Y$ axis, its $x$ co-ordinate is zero.

Here $x_1=-4,y_1=6$,

$x_2=8,y_2=-3$

$\therefore$ By section formula, $x=(m{x}_2-n{x}_1)/(m+n)$

$\therefore 0=(m\times 8+n\times (-4\left)\right)/(m+n)$

$\therefore 0=(8m+(-4n\left)\right)/(m+n)$

$\therefore 0=8m+(-4n)$

$\therefore 8m=4n$

$\therefore m/n=4/8=1/2$

Hence the ratio $m:n$ is $1:2$.

(ii) By Section formula $y=(m{y}_2+n{y}_1)/(m+n)$

Substitute $m$ and $n$ in above equation

$\therefore y=(1\ast -3+2\ast 6)/(1+2)$

$\therefore y=(-3+12)/3$

$\therefore y=9/3=3$

So the co-ordinates of the point of intersection are $(0,3)$.

(iii) By distance formula, $d\left(AB\right)=\sqrt{\left[\right(x_2-x_1{)}^2+(y_2-y_1{)}^2]}$

$\therefore d\left(AB\right)=\sqrt{\left[\right(8-(-4){)}^2+(-3-6{)}^2]}$

$\therefore d\left(AB\right)=\sqrt{\left[\right(12{)}^2+(-9{)}^2]}$

$\therefore d\left(AB\right)=\sqrt{(144+81)}$

$\therefore d\left(AB\right)=\sqrt{225}$

$\therefore d\left(AB\right)=15$.

Hence the length of $AB$ is $15$ units.