Question

find the ratio in which the line x-3y=0 divides the line segment joining the points -2 -5 and 6 3 find the coordinates of the the point of intersection

Answer 1

I have solved the problem please check it with the book back answer. If you have any questions please let me know.

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Answer 2

Answer:

The point of intersection is $(\dfrac{9}{2},\dfrac{3}{2})$

Step-by-step explanation:

Point A(-2,-5) and B(6,3) divides the line x-3y=0 in k:1 ratio at point P

Formula: Using section formula

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

$P(x,y)=(\dfrac{6k-2}{k+1},\dfrac{3k-5}{k+1})$

$x\rightarrow \dfrac{6k-2}{k+1}$

$y\rightarrow \dfrac{3k-5}{k+1}$

Substitute x and y into equation of line x-3y=0 and solve for k

$\dfrac{6k-2}{k+1}-3\cdot \dfrac{3k-5}{k+1}=0$

$6k-2-9k+15=0$

$-3k+13=0$

$k=\dfrac{13}{3}$

Point P divide AB in 13:3 ratio.

Put k=13/3 into $x\rightarrow \dfrac{6k-2}{k+1}$ and $y\rightarrow \dfrac{3k-5}{k+1}$

$x\rightarrow \dfrac{26-2}{13/3+1}\Rightarrow \dfrac{9}{2}$

$y\rightarrow \dfrac{13-5}{13/3+1}\Rightarrow \dfrac{3}{2}$

Hence, The point of intersection is $(\dfrac{9}{2},\dfrac{3}{2})$