The point which divides the line segment joining the points $ (7,-6) $ and $ (3,4) $ in ratio $ 1: 2 $ internally lies in the
(A) I quadrant
(B) II quadrant
(C) III quadrant
(D) IV quadrant

Given: 

A point which divides the line segment joining the points $(7,\ –6)$ and $(3,\ 4)$ in ratio $1 : 2$ internally.

To do: 

We have to find the quadrant in which the point lies in.

Solution:

Here $x_1=7,\ y_1=-6,\ x_2=3,\ y_2=4,\ m=1$ and $n=2$.

Using the division formula,

$( x,\ y)=( \frac{mx_2+nx_1}{m+n},\ \frac{my_2+ny_1}{m+n})$

$( x,\ y)=( \frac{1\times3+2\times7}{1+2},\ \frac{1\times4+2\times-6}{1+2})$

$( x,\ y)=( \frac{17}{3},\ \frac{-8}{3})$ which lies in IV quadrant as the x-coordinate is positive and the y-coordinate is negative.

Thus, $( \frac{17}{3},\ \frac{-8}{3})$ divides the line segment joining the points $(7,\ –6)$ and $(3,\ 4)$ in ratio $1 : 2$ internally and it lies in IV quadrant.

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Updated on: 10-Oct-2022

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