Find the ratio in which the line joining (-5,3) and (1,-3) divides the line joining (3,4) and (7,8)?

Question

the answer is given as 5:9 but my answer is coming negative!

Answer 1

Line joining (-5, 1) & (1, -3)

Slope - point form is (y - y ₁) = m(x - x ₁) where m = (y₂ - y₁)/(x₂ - x₁)

Slope m = (-3-1)/(1-(-5)) = -4/6 = -2/3

The line joining is (y - 1) = (-2/3)(x - (-5))

Multiply each side by 3.

3(y - 1) = (-2)(x + 5)

3y - 3 = -2x - 10.

Add 2x to each side.

2x + 3y - 3 = -10

Add 3 to each side.

2x + 3y = -7---------------->(1)

The line joining is (3, 4) & (7, 8)

Slope m = (8 - 4)/(7 - 3) = 4/4 = 1.

Therefore (y - 4) = (1)(x - 3)

(y - 4) = (x - 3)

Subtract y from each side.

- 4 = x - y - 3.

Add 3 to each side.

x - y = - 1----------------------->(2).

Multiply equation(2) by 2 and Subtract  (1) & (2).

2x + 3y = -7

2x -2 y = - 2

-----------------------

5y = -5

Divide each side by 5.

y = -1

Subtution y = -1 in the equation (2).

x - (-1) = - 1

x + 1 = -1

Subtract 1 from each side.

x = -2

Therefore point of intersection is  (-2, -1)

Distance of ( -2 ,-1) from ( -5,1) p₁= √[9 +4] = √13

Distance of ( -2,-1) from ( 1, -3) p₂ = √[9+ 4] = √13

So ratio is 1: 1

Distance of ( -2,-1) from (3,4) p3= √[25+25] = 5√2

Distance of (-2,-1) from ( 7,8) p4= √[81+81] = 9√2

So ratio is 5: 9 answer