# If the line segment joining the points (3, −4),

Question:

If the line segment joining the points $(3,-4)$, and $(1,2)$ is trisected at points $P(a,-2)$ and $Q\left(\frac{5}{3}, b\right)$, Then,

(a) $a=\frac{8}{3}, b=\frac{2}{3}$

(b) $a=\frac{7}{3}, b=0$

(c) $a=\frac{1}{3}, b=1$

(d) $a=\frac{2}{3}, b=\frac{1}{3}$

Solution:

We have two points $A(3,-4)$ and $B(1,2)$. There are two points $P(a,-2)$ and $Q\left(\frac{5}{3}, b\right)$ which trisect the line segment joining $A$ and $B$.

Now according to the section formula if any point $P$ divides a line segment joining $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ in the ratio $m$ : $n$ internally than,

$\mathrm{P}(x, y)=\left(\frac{m_{1}+m x_{2}}{m+n}, \frac{m y_{1}+m y_{2}}{m+n}\right)$

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

Now we will use section formula to find the co-ordinates of unknown point A as,

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

$=\left(\frac{7}{3},-2\right)$

Equate the individual terms on both the sides. We get,

$a=\frac{7}{3}$

Similarly, the point Q is the point of trisection of the line segment AB. So, Q divides AB in the ratio 2: 1

Now we will use section formula to find the co-ordinates of unknown point A as,

$Q\left(\frac{5}{3}, b\right)=\left(\frac{2(1)+1(3)}{1+2}, \frac{2(2)+1(-4)}{1+2}\right)$

$=\left(\frac{5}{3}, 0\right)$

Equate the individual terms on both the sides. We get,

$b=0$