Question

P divides distance between A(-2,1) and B(1,4) in the ratio 2:1 calculate the coordinates of the point P​

Answer 1

$Let \:P(x,y) \:divide \: AB \: internally \: in \: the \\ratio \: m : n = 2 : 1$

$\underline { \pink {Using \: the \: section \: formula :}}$

$P(x,y) = \Big( \frac{mx_{2}+nx_{1}}{m+n} , \frac{my_{2}+ny_{1}}{m+n}\Big)$

$A(-2,1) = ( x_{1}, y_{1}) \:and \: B(1,4) = ( x_{2} , y_{2} )$

$P = \Big( \frac{2\times 1 + 1\times (-2)}{2+1} , \frac{2\times 4 + 1\times 1}{2+1}\Big) \\= \Big( \frac{2-2}{3} , \frac{8+1}{3} \Big) \\= \Big( 0, \frac{9}{3}\Big) \\= ( 0, 3)$

Therefore.,

$\red{ Coordinates \:of \: point \: P } \green { = ( 0,3) }$

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Answer 2

Given that ,

• The point P divide AB in the ratio 2 : 1

• The two points are A(-2,1) and B(1,4)

Let , the cordinate of point P be P(x,y)

We know that , the section formula is given by

$\large \sf \fbox{(x,y)= \bigg( \frac{mx_{2}+nx_{1}}{m+n} , \frac{my_{2}+ny_{1}}{m+n}\bigg) \: }$

Thus ,

$\sf \mapsto P(x,y)= \bigg( \frac{2\times 1 + 1\times (-2)}{2+1} , \frac{2\times 4 + 1\times 1}{2+1}\bigg) \\ \\ \sf \mapsto P(x,y)= \bigg( \frac{2-2}{3} , \frac{8+1}{3} \bigg) \\ \\ \sf \mapsto P(x,y)= \bigg( 0, \frac{9}{3}\bigg) \\ \\ \sf \mapsto P(x,y)= ( 0, 3)$

$\sf \therefore \underline{The \: coordinate \: of \: point \: P \: is \: (0,3) \: }$