Question
Find the distance between the pair of points (-5, 7), (-1, 3)
✓
Answer
The points of trisection means that the points which divide the line into three equal parts. From the figure, it is clear that $C$, and $D$ are these two points. Let $C\left(x_1, y_1\right)$ and $D\left(x_2, y_2\right)$ are the points of trisection of the line segment joining the given points i.e., $B C=C D=D A$
Let $B C=C D=D A=k$, Point $C$ divides $B C$ and $C A$ as: $B C=k C A=C D+D A=k+k=2 k$ Hence the ratio between $B C$ and $C A$ is: $\frac{B C}{C A}=\frac{k}{2 k}=\frac{1}{2}$
Therefore, point $C$ divides $B A$ internally in the ratio 1:2 then by section formula we have that if a point $P(x, y)$ divides two points $P\left(x_1, y_1\right)$ and $Q\left(x_2, y_2\right)$ in the ratio $m: n$ then, the point $(x, y)$ is given by $(x, y)=\left(\frac{ mx _2+ mx _1}{m+ n }, \frac{ my _2+ ny _1}{m+ n }\right)$
Therefore $C ( x , y )$ divides $B (-2,-3)$ and $A (4,-1)$ in the ratio $1: 2$, then
$C(x, y)=\left(\frac{(1 \times 4)+(2 \times-2)}{1+2}, \frac{(1 \times-1)+(2 \times-3)}{1+2}\right)$
$C(x, y)=\left(\frac{4-4}{1+2}, \frac{-1-6}{1+2}\right)$
$C(x, y)=\left(0, \frac{-7}{3}\right)$
Point D divides the $B D$ and $D A$ as:DA $=k B D=B C+C D=k+k=2 k$
Hence the ratio between BD and DA is: $\frac{B D}{D A}=\frac{2 k}{k}=\frac{2}{1}$
The point $D$ divides the line $B A$ in the ratio 2:1
So now applying section formula again we get,
$D(x, y)=\left(\frac{(2 \times 4)+(1 \times-2)}{2+1}, \frac{(2 \times-1)+(1 \times-3)}{2+1}\right)$
$D(x, y)=\left(\frac{8-2}{3}, \frac{-2-3}{3}\right)$
$D(x, y)=\left(\frac{6}{3}, \frac{-5}{3}\right)$
$D(x, y)=\left(2, \frac{-5}{3}\right)$
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