Example :
In what ratio does the line x-y-2 = 0 divide the line segment joining the points A (3, -1) and B (8, 9)?
Solution :
Let the line x-y-2 = 0 is divided by the line segment joining the points A(3, -1) and B(8, 9) be a : 1
Using section formula internally :
= (lx2+mx1)/(l+m), (ly2+my1)/(l+m)
= a(8)+1(3)/(a+1), a(9)+1(-1)/(a+1)
= (8a+3)/(a+1), (9a-1)/(a+1)
The point (8a+3)/(a+1), (9a-1)/(a+1) lies on the line.
[(8a+3)/(a+1)]-[(9a-1)/(a+1)]-2 = 0
(8a+3)-(9a-1)-2(a+1) = 0
8a-9a+3+1-2a-2 = 0
-3a+2 = 0
3a = 2
a = 2/3
a : 1 ==> (2/3) : 1
So, the required ratio is 2 : 3.
Another method :
Equation of the line joining points A (3, -1) and B (8, 9)
(y-y1)/(y2-y1) = (x-x1)/(x2-x1)
(y+1)/(9+1) = (x-3)/(8-3)
(y+1)/10 = (x-3)/5
(y+1)/2 = (x-3)
y+1 = 2(x-3)
y+1 = 2x-6
2x-y-6-1 = 0
2x-y-7 = 0
Point of intersection of the lines 2x-y-7 = 0 and x-y-2 = 0
2x-y-7 = 0 -------(1)
x-y-2 = 0 -------(2)
(1)-(2)
2x-x-7+2 = 0
x-5 = 0
x = 5
By applying the value of x in (2), we get
5-y-2 = 0
3-y = 0
y = 3
So, both are intersecting at the point (5, 3).
Now, we find in what ratio the line joining two points A (3, -1) and B (8, 9) is being divided by the point (5, 3).
(8l+3m)/(l+m), (9l-1m)/(l+m) = (5, 3)
Equating x coordinates, we get
(8l+3m)/(l+m) = 5
8l+3m = 5l+5m
3l = 2m
l/m = 2/3
l:m = 2 : 3
So, the required ratio is 2 : 3.
Note : By equating y-coordinate also, we will get the same answer.
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