Question

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

Answer 1

let t line x-y-2=0 divid t line segment in t ratio k : 1 at point C

so coordinates of C are

{ ( 8k +3 / k+1) , ( 9k-1 / k+1 ) }

C lies on x - y - 2 = o

so, subs coordinates of x,y in tis eq.

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

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

-3k +2 / k+1 = 0

-3k + 2 = 0

=> k = 2/3

i.e. 2 : 3

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