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  • Find the ratio in which x-3y=0 divides the line segment joining the points A( -2,-5) and B (6,3) and also find the coordinates of the point .

Find the ratio in which x-3y=0 divides the line segment joining the points A( -2,-5) and B (6,3) and also find the coordinates of the point .

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Let the point $P$ divide the line segment joining $A(-2,-5)$ and $B(6,3)$ in the ratio $m:n$.

The equation of the line is: $x - 3y = 0$

Using the section formula, the coordinates of $P$ are: $\left(\dfrac{m(6)+n(-2)}{m+n},\dfrac{m(3)+n(-5)}{m+n}\right)$

So, $x = \dfrac{6m-2n}{m+n}$

$y = \dfrac{3m-5n}{m+n}$

Since $P$ lies on the line $x-3y=0$,

$\dfrac{6m-2n}{m+n}-3\left(\dfrac{3m-5n}{m+n}\right)=0$

Taking LCM: $6m-2n-9m+15n=0$

$-3m+13n=0$

$3m=13n$

$\dfrac{m}{n}=\dfrac{13}{3}$

Hence, the line divides the segment internally in the ratio: $13:3$

Now find the coordinates of the point using the section formula:

$P=\left(\dfrac{13(6)+3(-2)}{13+3},\dfrac{13(3)+3(-5)}{13+3}\right)$

$=\left(\dfrac{78-6}{16},\dfrac{39-15}{16}\right)$

$=\left(\dfrac{72}{16},\dfrac{24}{16}\right)$

$=\left(\dfrac{9}{2},\dfrac{3}{2}\right)$

Therefore, the required point is $\left(\dfrac{9}{2},\dfrac{3}{2}\right)$.

Posted by

Sanskriti Srivastava

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