The point $P\left(\dfrac{3}{4},\dfrac{5}{12}\right)$ divides the line segment joining points $A\left(\dfrac{1}{2},\dfrac{3}{2}\right)$ and $B(2,-5)$ .

To find:

The ratio in which he point P divides the segment AB.

Solution:

Section formula: If a point divides a segment in m:n, then the coordinates of that point are,

$Point=\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)$

Let point P divides the segment AB in m:n. Then by using the section formula, we get

$\left(\dfrac{3}{4},\dfrac{5}{12}\right)=\left(\dfrac{m(2)+n(\dfrac{1}{2})}{m+n},\dfrac{m(-5)+n(\dfrac{3}{2})}{m+n}\right)$

$\left(\dfrac{3}{4},\dfrac{5}{12}\right)=\left(\dfrac{2m+\dfrac{n}{2}}{m+n},\dfrac{-5m+\dfrac{3n}{2}}{m+n}\right)$

On comparing both sides, we get

$\dfrac{3}{4}=\dfrac{2m+\dfrac{n}{2}}{m+n}$

$\dfrac{3}{4}(m+n)=\dfrac{4m+n}{2}$

Multiply both sides by 4.

$3(m+n)=2(4m+n)$

$3m+3n=8m+2n$

$3n-2n=8m-3m$

$n=5m$

It can be written as

$\dfrac{1}{5}=\dfrac{m}{n}$

$1:5=m:n$

Therefore, the point P divides the line segment AB in 1:5.

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3 years ago