Question

find the ratio in which line 3x+y-9 divids line segment joining a (1,3) and b (2,7)

Answer 1

• Given , a line 3x+y-9 = 0

• R.T.P: ratio in which it divides the ine segment joining a (1,3) and b (2,7)
• From section formula,   we know that it tells us coordinates of the point which divides the given line segment into two parts so that their lengths may be in the ratio m:n
• If a point P(x,y) divides line joining into 2 parts A=(a,b) and B=(c,d) in ratio m:n internally, then the coordinates are given by,
• (x,y) = (c.m+a.n/(m+n),  d.m+b.n/(m+n))
• By using above formula, we get,
• x=2k+1/k+1 and y=7k+3/k+1
• Put x,y values in the given line 3x+y-9=0

⇒3((2k+1)/(k+1)) + ((7k+3)/k+1)-9=0

⇒6k+3+7k+3=9(k+1)

⇒4k-3=0

• k=4/3

• Hence,the ratio in which line 3x+y-9=0 divides line segment joining a (1,3) and b (2,7) is 3:4