Question

Find the coordinate of points which trisect the line segment joining $(1,-2)$ and $(-3,4)$.

Answer 1

Let $C$ and $D$ be the points of trisection.

Then, $AC=CD=DB$.

Let, $A(1,-2)=A(x_1,y_1)$

And, $B(-3,4)=B(x_2,y_2)$

Now, $C$ divides $AB$ in the ratio $1:2$ and $D$ divides $AB$ in the ratio $2:1$.

Therefore,

Co-ordinates of $C$ using section formula,

$=[\frac{m\times x_2+n\times x_1}{m+n},\frac{m\times y_2+n\times y_1}{m+n}]$

$=(\frac{1\times (-3)+2\times \left(1\right)}{2+1},\frac{1\times \left(4\right)+2\times (-2)}{2+1})$

$=(-\frac{1}{3},0)$

And co-ordinates of $D$,

$=(\frac{2\times (-3)+1\times \left(1\right)}{2+1},\frac{2\times \left(4\right)+1\times (-2)}{2+1})$

$=(-\frac{5}{3},2)$

Hence, the two points of trisection are $C(-\frac{1}{3},0),D(-\frac{4}{3},2)$.