Question

Find the ratio in which the line joining (-5,1) and (1,-3) divides the line joining (3,4) and (7,8). Also find the coordinates of the point of intersection.​

Answer 1

Answer:

-5:9

Step-by-step explanation:

it is a question of extrrnal intersection.

Answer 2

Answer:

The ratio dividing the line joining (3,4) and (7,8) is -5:9 and the point of intersection of two lines is (-2, -1).

Step-by-step explanation:

The equation of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is

 $\dfrac{y-y_1}{x - x_1} = \dfrac{y_2-y_1}{x_2-x_1}$

Let $A(-5,1)$ and $B(1,-3)$, then the line AB can be written as

$\dfrac{y-1}{x - (-5)} = \dfrac{-3-1}{1-(-5)}$

$\dfrac{y-1}{x +5} = \dfrac{-4}{6}$

$3(y-1} ) = {-2}({x +5})$

$3y-3 = -2x -10$

$3y-3 +2x +10=0 \implies 3y+2x +7=0$                                   ...(1)

Let $C(3,4)$ and $D(7,8)$, then the line CD can be written as

$\dfrac{y-4}{x -3} = \dfrac{8-4}{7-3}$

$\dfrac{y-4}{x -3} = \dfrac{4}{4}$

$y-4= x -3$

$y-4- x+3=0\implies y- x-1=0$                                           ...(2)

Multiplying equation (2) and adding to equation (1),

$3y+2x +7+2(y- x-1)=0$

$3y+2x +7+2y- 2x-2=0$

$5y +5=0\implies y=-1$

Substituting y in equation (2),  

$-1- x-1=0\implies x=-2$

Therefore, the co-ordinates of the point of intersection is (-2, -1).

The ratio m : n that divides the line joining $(x_1, y_1)$ and $(x_2, y_2)$ is

 $\dfrac{mx_2+nx_1}{m+n} , \dfrac{my_2+ny_1}{m+n}$

The co-ordinates (-2, -1) divides the line joining (3,4) and (7,8) in m:n ratio. $\therefore \dfrac{m\times 7+n\times 3}{m+n}=-2, \dfrac{m\times 8+n\times 4}{m+n}=-1$

Consider the first co-ordinate,

$\dfrac{7m+3n}{m+n}=-2$

${7m+3n}=-2(m+n)$

$7m+3n=-2m-2n$

$9m=-5n \implies \dfrac{m}{n} =\dfrac{-5}{9}$

Therefore, the ratio dividing the line joining (3,4) and (7,8) is -5:9.