Let P and Q be the points of trisection of the line segment joining the points $A( 2,\ -2)$ and $B( -7,\ 4)$ such that P is nearer to A. Find the coordinates of P and Q.
Given: P and Q be the points of trisection of the line segment joining the points $A( 2,\ -2)$ and $B( -7,\ 4)$ such that P is nearer to
To do: To find the coordinates of P and Q.
Solution:
Since P and Q are the points of trisection of AB,
$\therefore AP =PQ = QB$
Thus, P divides AB internally in the ratio 1 : 2
And Q divides AB internally in the ratio 2:1.
$\therefore$ Using section formula, we have$( x,\ y) =(\frac{nx_{1} +mx_{2}}{m+n} ,\ \frac{ny_{1} +my_{2}}{m+n})$
For P ,$m=1$ and $n=2$
$\therefore P( x,\ y) =( \frac{2\times 2+1\times -7}{1+2} ,\ \frac{2\times -2+1\times 4}{1+2})$
$ =(\frac{4-7}{3} ,\ \frac{-4+4}{3})$
$=(\frac{-3}{3} ,\ 0)$
$=( -1,\ 0)$
And for Q, $m=2$ and $n=1$
$\therefore Q=(\frac{1\times -7+2\times 2}{1+2} ,\ \frac{1\times -2+2\times 4}{1+2})$
$=(\frac{-7+4}{3} ,\ \frac{-2+8}{3})$
$=( -1,\ 2)$
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