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 point of intersection.

Answer 1

Let the given line $ax+by+c=0$ divide the line segment joining $A(x_1,y_1)$ and $B(x_2,y_2)$ in the ratio m : n, then

$\frac{m}{n}=-\frac{ax_1+by_1+c}{ax_2+by_2+c}$

solution : Let line x-3y=0x−3y=0 divide the line segment (-2,-5) and (6,3) in the ratio of m : n

$\frac{m}{n}=-\frac{-2-3(-5)}{6-3(3)}\\\\=-\frac{-2+15}{6-9}\\\\=-\frac{13}{-3}$

hence, m : n = 13 : 3

here positive sign shows that A and B are on opposite sides of the given line.

Let (x, y) is the coordinates of point of intersection.

use section formula,

$x=\frac{mx_2+nx_1}{m+n}\\\\=\frac{13\times6+3\times-2}{13+3}=\frac{78-6}{16}=\frac{72}{16}=4.5$

$y=\frac{my_2+ny_1}{m+n}\\\\=\frac{13\times3+3\times-5}{13+3}=\frac{39-15}{16}=\frac{3}{2}=1.5$

hence, (4.5,1.5) is the point of intersection.

Answer 2

Answer:

hence, (4.5,1.5) is the point of intersection.

Step-by-step explanation:

You can find this out by using the section formula.

just substitute the values of x₁, x₂, y₁ and y₂ in the section formula which is:

[x,y] = ([m₁x₂+m₂x₁/m₁+m₂) , (m₁y₂+m₂y₁/m₁+m₂)]