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A(20, 0) and B (10, -20) are two fixed points. Find the co-ordinates of the point P in AB such that : 3PB = AB. Also, find the co-ordinates of some other point Q in AB such that AB = 6AQ.

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Answer

From figure,

A(20, 0) and B (10, -20) are two fixed points. Find the co-ordinates of the point P in AB such that : 3PB = AB. Also, find the co-ordinates of some other point Q in AB such that AB = 6AQ. Section and Mid-Point Formula, Concise Mathematics Solutions ICSE Class 10.

⇒ AB = AP + PB

⇒ 3PB = AP + PB

⇒ AP = 2PB

APPB=21\dfrac{AP}{PB} = \dfrac{2}{1}

⇒ AP : PB = 2 : 1.

Let the co-ordinates of P be (x, y)

x=m1x2+m2x1m1+m2=2×10+1×202+1=20+203=403.\therefore x = \dfrac{m1x2 + m2x1}{m1 + m2} \\[1em] = \dfrac{2 \times 10 + 1 \times 20}{2 + 1} \\[1em] = \dfrac{20 + 20}{3} \\[1em] = \dfrac{40}{3}.

and,

y=m1y2+m2y1m1+m2=2×20+1×02+1=40+03=403.y = \dfrac{m1y2 + m2y1}{m1 + m2} \\[1em] = \dfrac{2 \times -20 + 1 \times 0}{2 + 1} \\[1em] = \dfrac{-40 + 0}{3} \\[1em] = \dfrac{-40}{3}.

From figure,

⇒ AB = AQ + QB

⇒ 6AQ = AQ + QB

⇒ QB = 5AQ

AQQB=15\dfrac{AQ}{QB} = \dfrac{1}{5}

⇒ AQ : QB = 1 : 5.

Let the co-ordinates of Q be (p, q)

p=m1x2+m2x1m1+m2=1×10+5×201+5=10+1006=1106=553.\therefore p = \dfrac{m1x2 + m2x1}{m1 + m2} \\[1em] = \dfrac{1 \times 10 + 5 \times 20}{1 + 5} \\[1em] = \dfrac{10 + 100}{6} \\[1em] = \dfrac{110}{6} = \dfrac{55}{3}.

and,

q=m1y2+m2y1m1+m2=1×20+5×01+5=20+06=103.q = \dfrac{m1y2 + m2y1}{m1 + m2} \\[1em] = \dfrac{1 \times -20 + 5 \times 0}{1 + 5} \\[1em] = \dfrac{-20 + 0}{6} \\[1em] = \dfrac{-10}{3}.

Hence, P = (403,403) and Q=(553,103)\Big(\dfrac{40}{3}, -\dfrac{40}{3}\Big) \text{ and Q} = \Big(\dfrac{55}{3}, -\dfrac{10}{3}\Big).

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