Question

If P and Q are the points of trisection of the line joining A(2, -2) and B(-7, 4) such that P is nearer to A, then the coordinates of P and Q are:

• A. P(0, -1), Q(-4, 2)
• B. P(-1, 0), Q(-4, 2)
• C. P(-1, 0), Q(4, -2)
• D. P(1, 0), Q(-4, -2)

Answer 1

The correct option is B P(-1, 0), Q(-4, 2)

Given:P and Q are the points of trisection of the line joining A(2, -2) and B(-7, 4) such that P is nearer to A.
Let the coordinates of P and Q be $P(x_1,y_1)andQ(x_2,y_2)$.

$\because$ Point P is near to A.

$\therefore$ $P(x_1,y_1)$ divides AB in the ratio 1 : 2, while $Q(x_2,y_2)$ divides AB in the ratio 2 : 1.
section formula;

$P(x,y)=\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2}$
Thus, by section formula,
$P(x_1,y_1)=\frac{(-7+4)}{3},\frac{(4-4)}{3}=(-1,0)$
$Q(x_2,y_2)=\frac{(-14+2)}{3},\frac{(8-2)}{3}=(-4,2)$

Hence, coordinates of point P is $(-1,0)$ and point Q is $(-4,2)$.