Question

In what ratio does the line x-y-2=0 divide the line segment joining the point A(3,-1) and B(8,9)

Answer 1

Let(α,β) cuts AB in k:1
A(3,-1) B(8,9)
using section formula,
α=(8k+3)/(k+1)
β=(9k-1)/(k+1)
put α,β in given equation
8k+3-9k+1-2(k+1)=0
-3k+2=0
k=3/2
So ratio is 3:2internally

Answer 2

Hey there!

Question:

In what ratio does the line x-y-2=0 divide the line segment joining the points A (3, -1) and B (8, 9)?

Answer:

Let the ratio be k:1.

And, point of intersection be (X,Y).

Now,

X = (m₁x₂ + m₂x₁) /( m₁ + m₂)

X = (k*8 + 1*3) / (k+1)

X = (8k + 3) /( k + 1)

And,

Y =  (m₁y₂ + m₂y₁) /( m₁ + m₂)

Y = ( k* 9 - 1*1) / (k+1)

Y = (9k-1)/(k+1)

Now,

Since, the point (X,Y) also lies on x - y - 2 = 0.

So, it will satisfy the given equation,

x - y - 2 = 0.

(8k+3)/(k+1) - (9k-1)/(k+1) - 2 = 0

⇒ (8k + 3 - 9k - 1)/ (k+1) = 2

⇒ -k +4 = 2k + 2

⇒ 4 = 2k+ k + 2

⇒2  = 3k

⇒ 2/3 = k

or, k = 2 / 3

Now,

Ratio = k : 1 = (2 / 3) : 1 = 2 : 3