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 coordinated of the point of intersection.​

Answer 1

Let the point of intersection be P (x,y) and divide AB {A(-2,-5) , B (6,3)} in the ratio k:1

Then use the section formula to find the coordinates of P.

Next put ghe values of x and y in the equation and you will get the ratio.

Then put it and find x and y.
The answer is:- Ratio 13:3 and coordinates of intersection are (9/2,3/2)

Easy!

Regards

Wolfwithscythe

Answer 2

Explanation:

Given that

The given line is x-3y = 0……… (1)

Let A = (-2,-5) and B = (6, 3)

Equation of AB is   ((y-y1))/((x-x1))  =  ((y2-y1))/((x2-x1))  

((y+5))/((x+2))  =  ((3+5))/((6+2))

((y+5))/((x+2))=8/8

((y+5))/((x+2) )=1

x+2=y+5

x-y=3  ----- (2)

Solving (1) and (2) we will get x =9/2   and y=3/2

So the point of intersection is P (9/2, 3/2).

Let P divide AB in the ratio k:1

The applying the section formula we get

(9/2,3/2)={((k6+1(-2)))/((k+1) ),((k3+1(-5)))/((k+1) )}  

(9/2,3/2)={((6k-2))/((k+1) ),((3k-5))/((k+1) )}

Equating y-coordinates

3/2=(3k-5)/(k+1)

6k-10 = 3k+3

6k-3k=3+10

3k=13

k=13/3

Hence the required ratio is 13/3:1 or 13:3

To read more:

https://brainly.in/question/8632438

https://brainly.in/question/8619266