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ICSE Class 10 Mathematics Question 18 of 21

Solved 2024 Specimen Paper ICSE Class 10 Mathematics — Question 14

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14
Question

Question 8(iii)

A(a, b), B(-4, 3) and C(8, -6) are the vertices of a △ ABC. Point D is on BC such that BD : DC is 2 : 1 and M(6, 0) is mid-point of AD. Find :

(a) coordinates of point D.

(b) coordinates of point A.

(c) equation of a line parallel to line BC, through M.

Answer

(a) Given,

BD : DC = 2 : 1.

A(a, b), B(-4, 3) and C(8, -6) are the vertices of a △ ABC. Point D is on BC such that BD : DC is 2 : 1 and M(6, 0) is mid-point of AD. Find : ICSE 2024 Maths Specimen Solved Question Paper.

Let coordinates of D be (x, y)

By section-formula,

(x, y) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting values, we get :

⇒(x,y)=(2×8+1×(−4)2+1,2×−6+1×32+1)⇒(x,y)=(16−43,−12+33)⇒(x,y)=(123,−93)⇒(x,y)=(4,−3).\Rightarrow (x, y) = \Big(\dfrac{2 \times 8 + 1 \times (-4)}{2 + 1}, \dfrac{2 \times -6 + 1 \times 3}{2 + 1}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{16 - 4}{3}, \dfrac{-12 + 3}{3}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{12}{3}, \dfrac{-9}{3}\Big) \\[1em] \Rightarrow (x, y) = (4, -3).

Hence, coordinates of D = (4, -3).

(b) By mid-point formula,

Mid-point = (x1+x22,y1+y22)\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

Given,

M(6, 0) is the mid-point of AD.

∴(6,0)=(a+42,b+(−3)2)⇒(6,0)=(a+42,b−32)⇒a+42=6 and b−32=0⇒a+4=12 and b−3=0⇒a=12−4=8 and b=3.\therefore (6, 0) = \Big(\dfrac{a + 4}{2}, \dfrac{b + (-3)}{2}\Big) \\[1em] \Rightarrow (6, 0) = \Big(\dfrac{a + 4}{2}, \dfrac{b - 3}{2}\Big) \\[1em] \Rightarrow \dfrac{a + 4}{2} = 6 \text{ and } \dfrac{b - 3}{2} = 0 \\[1em] \Rightarrow a + 4 = 12 \text{ and } b - 3 = 0 \\[1em] \Rightarrow a = 12 - 4 = 8 \text{ and } b = 3.

A = (a, b) = (8, 3).

Hence, coordinates of A = (8, 3).

(c) By formula,

Slope of line = y2−y1x2−x1\dfrac{y_2 - y_1}{x_2 - x_1}

Slope of line BC = −6−38−(−4)=−912=−34\dfrac{-6 - 3}{8 - (-4)} = \dfrac{-9}{12} = -\dfrac{3}{4}.

We know that,

Slope of parallel lines are equal.

Slope of line parallel to BC = −34-\dfrac{3}{4}.

By point-slope form,

Equation of line : y - y1 = m(x - x1)

Equation of line parallel to BC and passing through M is :

⇒ y - 0 = −34(x−6)-\dfrac{3}{4}(x - 6)

⇒ y = −34(x−6)-\dfrac{3}{4}(x - 6)

⇒ 4y = -3(x - 6)

⇒ 4y = -3x + 18

⇒ 3x + 4y = 18.

Hence, equation of the required line is 3x + 4y = 18.