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Question

The point A divides the line segment joining the points P(5,1) and Q(3,5) in the ratio k:1. Find the two values of k for which the area of ABC is equal to 2 square units where B is (1,5) and C(7,2) .

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Solution

It is given that point A divides the join of P(5,1) and Q(3,5) in the ratio of K:1.

Here, x1=5,y1=1, x2=3,y2=5,m=k and n=1


P(X,Y)=(mx1+nx2m+n,my1+ny2m+n)

So, The Coordinate of A =(3k5k+1,5k+1k+1)


we have,
Area of ABC=2 Sq.units

12(3k5k+1×52+7×5k1k+1)(5k+1k+1×1+352×3k1k+1)=2

12(15k25k+12+5k1k+1)(5k+1k+1+356k10k+1)=2

12(15k252k2+35k+7)(5k+1+35k+356k+10)k+1=2

1214k66k+1=27k33k+1=±2

7k=33=±(k+1)

7k33=2k+2,7k33=2k2


5k=35,9k=31k=7,319


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