Question

let P and Q be the points of trisection of the line segment joining the points A (2, - 2)and B( - 7,4) such that P is nearer to find the coordinates of P and Q

Answer 1

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Answer 2

Answer:

P ≡ (-1,0) and Q ≡ (-4,2)

Step-by-step explanation:

P and Q are the points of trisection of AB, therefore AP = PQ = QB .Thus, P divides AB internally in ratio 1:2 and Q divides AB internally in the ratio 2:1.

So, By using section formula : If a point C divides the line segment having end points A(a',a'') and B(b',b'') internally in the ratio m:n Then coordinates of C are given by :

$C = (\frac{mb'+na'}{m+n} ,\frac{mb''+na''}{m+n})$

$\text{So, coordinates of P =}[\frac{1(-7)+2(2)}{1+2} ,\frac{1(4)+2(-2)}{1+2} ]\\\\=(\frac{-7+4}{3} ,\frac{4-4}{3}) \\\\=(\frac{-3}{3},0)\\\\=(-1,0)\\\\\text{So, coordinates of Q =}[\frac{2(-7)+1(2)}{2+1} ,\frac{2(4)+1(-2)}{2+1} ]\\\\=(\frac{-14+2}{3} ,\frac{8-2}{3}) \\\\=(\frac{-12}{3},\frac{6}{3})\\\\=(-4,2)$

Hence, P ≡ (-1,0) and Q ≡ (-4,2)