Given:
The point [tex]P\left(\dfrac{3}{4},\dfrac{5}{12}\right)[/tex] divides the line segment joining points [tex]A\left(\dfrac{1}{2},\dfrac{3}{2}\right)[/tex] and [tex]B(2,-5)[/tex].
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,
[tex]Point=\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)[/tex]
Let point P divides the segment AB in m:n. Then by using the section formula, we get
[tex]\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)[/tex]
[tex]\left(\dfrac{3}{4},\dfrac{5}{12}\right)=\left(\dfrac{2m+\dfrac{n}{2}}{m+n},\dfrac{-5m+\dfrac{3n}{2}}{m+n}\right)[/tex]
On comparing both sides, we get
[tex]\dfrac{3}{4}=\dfrac{2m+\dfrac{n}{2}}{m+n}[/tex]
[tex]\dfrac{3}{4}(m+n)=\dfrac{4m+n}{2}[/tex]
Multiply both sides by 4.
[tex]3(m+n)=2(4m+n)[/tex]
[tex]3m+3n=8m+2n[/tex]
[tex]3n-2n=8m-3m[/tex]
[tex]n=5m[/tex]
It can be written as
[tex]\dfrac{1}{5}=\dfrac{m}{n}[/tex]
[tex]1:5=m:n[/tex]
Therefore, the point P divides the line segment AB in 1:5.