In ` triangle A B C ,` a point `P` is taken on `A B` such that `A P//B P=1//3` and point `Q` is taken on `B C` such that `C Q//B Q=3//1` . If `R` is the point of intersection of the lines `A Qa n dC P ,` using vector method, find the area of ` A B C` if the area of ` B R C` is 1 unit

Answer

`13/9` sq. unit

Explanation

Taking A as origin let `vecb and vecc` be the postion vectors of B and C , respectively,
The position vector of Q is `(3 vecb + vecc) / 4 ` and that of p is `vecb/ 4`
`if (AR)/(QR) = lambda/1` then position vector of R = ` lambda (( 3vecb +vecc)/4)` ltbRgt if `(CR)/(RP) = mu/1 ` then position vector of R = `(muvecb/4+vecc)/(mu+1)` (ii)
comparing (i) and (ii) , we have,
`(3lambda)/4 = mu/(4(mu+1))and lambda/4 = 1/(mu+1)`
solving , `lambda = 4/13 and mu = 12`
Therefore, position , vector R is `(3 vecb +vecc) / 13`
`triangleABC and triangleBRC` have the same base. there area are proporotional to AQ and RQ
`(triangleABC)/(triangleBRC)=|(3vecb +vecc)/4|/(|(3vecb +vecc)/4-((3vecb +vecc)/13)|)= 13/9`
Area of `triangleABC` is 13/9 units.