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Find the coordinates of the points which divide the line segment joining A $(-2, 2)$ and B $(2, 8)$ into four equal parts.
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Sol. Let points P, Q and R divide the line segment joining A $(-2, 2)$ and B $(2, 8)$ into four equal parts.
$\therefore$ P divides AB in the ratio $1:3$ or AP : PB = $1:3$
So, coordinates of P = $(\frac{1\times2+3\times(-2)}{1+3}, \frac{1\times8+3\times2}{1+3}) = (-1, \frac{7}{2})$
$\therefore$ Q divides AB in the ratio $1:1$ or AQ : QB = $1:1$
So, coordinates of P = $(\frac{1\times2+1\times(-2)}{1+1}, \frac{1\times8+1\times2}{1+1}) = (0,5)$
$\therefore$ R divides AB in the ratio $3:1$ or AR : RB = $3:1$
So, coordinates of P = $(\frac{3\times2+1\times(-2)}{3+1}, \frac{3\times8+1\times2}{3+1}) = (1, \frac{13}{2})$
$\therefore$ P divides AB in the ratio $1:3$ or AP : PB = $1:3$
So, coordinates of P = $(\frac{1\times2+3\times(-2)}{1+3}, \frac{1\times8+3\times2}{1+3}) = (-1, \frac{7}{2})$
$\therefore$ Q divides AB in the ratio $1:1$ or AQ : QB = $1:1$
So, coordinates of P = $(\frac{1\times2+1\times(-2)}{1+1}, \frac{1\times8+1\times2}{1+1}) = (0,5)$
$\therefore$ R divides AB in the ratio $3:1$ or AR : RB = $3:1$
So, coordinates of P = $(\frac{3\times2+1\times(-2)}{3+1}, \frac{3\times8+1\times2}{3+1}) = (1, \frac{13}{2})$