The point P(5, – 3) is one of the two points

True

Let P (5,-3) divides the line segment joining the points A (7,-2) and B (1 ,-5) in the ratio k: 1 internally.

By section formula, the coordinate of point P will be

$\left(\frac{k(1)+(1)(7)}{k+1}, \frac{k(-5)+1(-2)}{k+1}\right)$

i.e., $\left(\frac{k+7}{k+1}, \frac{-5 k-2}{k+1}\right)$

Now, $(5,-3)=\left(\frac{k+7}{k+1}, \frac{-5 k-2}{k+1}\right)$

$\Rightarrow$ $\frac{k+7}{k+1}=5$

$\Rightarrow \quad k+7=5 k+5$

$\Rightarrow$ $-4 k=-2$

$\therefore$ $k=\frac{1}{2}$

So the point P divides the line segment AB in ratio 1: 2. Hence, point P in the point of trisection of AB.