Question 2

Line Segment Joining: The line segment joining two points A and B is the straight path connecting them.

Trisection: Trisecting a line segment means dividing it into three equal parts. This gives two points — P and Q — such that $AP = PQ = QB$. So P divides AB in ratio $1:2$ and Q divides AB in ratio $2:1$.

Section Formula: If a point $\text{P}(x, y)$ divides the line segment joining $\text{A}(x_1, y_1)$ and $\text{B}(x_2, y_2)$ in the ratio $m:n$ internally, then:

$x = \frac{mx_2 + nx_1}{m + n}, \quad y = \frac{my_2 + ny_1}{m + n}$

We will use the section formula to find the points that divide the line segment.

Step 1 — First point of trisection

Let the given points be $A = (4, -1)$ and $B = (-2, -3)$. Let $P(x, y)$ be the first point of trisection. Point $P$ divides the line segment $AB$ in the ratio 1:2. We use the section formula: $x = \frac{m x_2 + n x_1}{m+n}$ and $y = \frac{m y_2 + n y_1}{m+n}$. Here, $(x_1, y_1) = (4, -1)$, $(x_2, y_2) = (-2, -3)$, $m=1$, $n=2$.

Let's find the x-coordinate of P. $x = \frac{1 \times (-2) + 2 \times 4}{1+2}$

$= \frac{-2 + 8}{3}$

$= \frac{6}{3}$

$= 2$

Now, let's find the y-coordinate of P. $y = \frac{1 \times (-3) + 2 \times (-1)}{1+2}$

$= \frac{-3 - 2}{3}$

$= \frac{-5}{3}$

$\boxed{P = \left(2, -\frac{5}{3}\right)}$

Step 2 — Second point of trisection

Let $Q(x, y)$ be the second point of trisection. Point $Q$ divides the line segment $AB$ in the ratio 2:1. We use the section formula again. Here, $(x_1, y_1) = (4, -1)$, $(x_2, y_2) = (-2, -3)$, $m=2$, $n=1$.

Let's find the x-coordinate of Q. $x = \frac{2 \times (-2) + 1 \times 4}{2+1}$

$= \frac{-4 + 4}{3}$

$= \frac{0}{3}$

$= 0$

Now, let's find the y-coordinate of Q. $y = \frac{2 \times (-3) + 1 \times (-1)}{2+1}$

$= \frac{-6 - 1}{3}$

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

$\boxed{Q = \left(0, -\frac{7}{3}\right)}$

Answer

(i) The first point of trisection is $\left(2, -\frac{5}{3}\right)$. (ii) The second point of trisection is $\left(0, -\frac{7}{3}\right)$.