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A(-8, 0), B(0, 16) and C(0, 0) are the vertices of a triangle ABC. Point P lies on AB and Q lies on AC such that AP : PB = 3 : 5 and AQ : QC = 3 : 5.

Show that : PQ = 38\dfrac{3}{8}BC.

Section Formula

Answer

The triangle is shown in the figure below:

A(-8, 0), B(0, 16) and C(0, 0) are the vertices of a triangle ABC. Point P lies on AB and Q lies on AC such that AP : PB = 3 : 5 and AQ : QC = 3 : 5. Section and Mid-Point Formula, Concise Mathematics Solutions ICSE Class 10.

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

x=m1x2+m2x1m1+m2=3×0+5×83+5=0408=408=5.\therefore x = \dfrac{m1x2 + m2x1}{m1 + m2} \\[1em] = \dfrac{3 \times 0 + 5 \times -8}{3 + 5} \\[1em] = \dfrac{0 - 40}{8} \\[1em] = \dfrac{-40}{8} = -5.

and,

y=m1y2+m2y1m1+m2=3×16+5×03+5=48+08=6.y = \dfrac{m1y2 + m2y1}{m1 + m2} \\[1em] = \dfrac{3 \times 16 + 5 \times 0}{3 + 5} \\[1em] = \dfrac{48 + 0}{8} \\[1em] = 6.

P = (x, y) = (-5, 6).

Let the co-ordinates of Q be (m, n)

m=m1x2+m2x1m1+m2=3×0+5×83+5=0408=408=5.\therefore m = \dfrac{m1x2 + m2x1}{m1 + m2} \\[1em] = \dfrac{3 \times 0 + 5 \times -8}{3 + 5} \\[1em] = \dfrac{0 - 40}{8} \\[1em] = \dfrac{-40}{8} = -5.

and,

n=m1y2+m2y1m1+m2=3×0+5×03+5=0+08=0.n = \dfrac{m1y2 + m2y1}{m1 + m2} \\[1em] = \dfrac{3 \times 0 + 5 \times 0}{3 + 5} \\[1em] = \dfrac{0 + 0}{8} \\[1em] = 0.

Q = (m, n) = (-5, 0).

Distance between two points = (x2x1)2+(y2y1)2\sqrt{(x2 - x1)^2 + (y2 - y1)^2}

PQ=(5(5))2+(06)2=02+(6)2=36=6 units.BC=(00)2+(016)2=02+(16)2=256=16 units.PQ = \sqrt{(-5 - (-5))^2 + (0 - 6)^2} \\[1em] = \sqrt{0^2 + (-6)^2} \\[1em] = \sqrt{36} \\[1em] = 6 \text{ units}. \\[1em] BC = \sqrt{(0 - 0)^2 + (0 - 16)^2} \\[1em] = \sqrt{0^2 + (-16)^2} \\[1em] = \sqrt{256} \\[1em] = 16 \text{ units}.

⇒ BC = 16 units

38BC=16×38\dfrac{3}{8}BC = 16 \times \dfrac{3}{8} = 6 units = PQ.

Hence, proved that PQ = 38\dfrac{3}{8}BC.

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