Question

Find the co-ordinates of a point which divides the line joining the points (-3,9) and (1,-3) internally in the ratio 2:3.

Answer 1

Answer :-

$\to\boxed{ \sf{P(x,y)=\bigg(\frac{-7}{5} \: ,\frac{21}{5}\bigg)}}$

To Find :-

→ Co-ordinates of a point that divides the line joining the given points .

Explanation :-

Given points are A(-3,9) and B(1,-3)

Let P(x,y) be a point that divides the line joing the point A and B in the ratio 2:3 internally .

As we know that ,

( By sectional formula ) → For internally

$\sf{if \: a \: point \: P(x,y) \: divides \: the \: line \:} \\ \sf{ joining \: points \: (x_{1},y_{1} ) \: and \: \: (x_{2},y_{2} ) \: } \\ \sf{ in \: the \: ratio \: m : n \: (internally) \: } \\ \\ \to \boxed{ \sf{P(x,y) =\bigg( \frac{mx_{2} + nx_{1}}{m + n} , \:\frac{my_{2} + ny_{1} }{m + n} \bigg)}}$

hence required point is given by -

$\to \sf{P= \bigg(\frac{ 2 \times 1 + 3 \times ( - 3)}{2 + 3} \: , \: \frac{2 \times ( - 3) +3 \times 9 }{2 + 3} \bigg)}\\ \\ \to \sf{ P= \bigg( \frac{2 - 9}{ 5} , \: \frac{ - 6 +27}{ 5} \bigg)} \\ \\ \to \boxed{ \sf{P (x,y)=\bigg(\frac{-7}{5}\: , \:\frac{21}{5}\bigg)}}$

Answer 2

Formula:

Let the point $P\:\:(x,\:y)$ divide the straight line joining the points $A\:\:(x_{1},\:y_{1})$ and $B\:\:(x_{2},\:y_{2})$ in the ratio $l : m$ internally. Then the coordinates of the point $P$ be

$\quad\quad\big(\frac{lx_{2}+mx_{1}}{l+m},\:\frac{ly_{2}+my_{1}}{l+m}\big)$

Solution:

Given straight line is made by joining the two points $(- 3,\:9)$ and $(1,\:- 3)$.

Then the coordinates of the point dividing the above straight line in the ratio $2 : 3$ internally are

$\quad\big(\frac{2(1)+3(-3)}{2+3},\:\frac{2(-3)+3(9)}{2+3}\big)$

$\to \big(\frac{2-9}{5},\:\frac{-6+27}{5}\big)$

$\to \big(\frac{-7}{5},\:\frac{21}{5}\big)$

$i.e.,\:(-\frac{7}{5},\:\frac{21}{5}\big)$

Answer:

$\quad$ The required point is $(-\frac{7}{5},\:\frac{21}{5})$.