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ICSE Class 10 Mathematics Question 17 of 21

Solved 2024 Question Paper ICSE Class 10 Mathematics — Question 11

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Question

Question 7(iii)

In the given diagram, ABC is a triangle, where B(4, -4) and C(-4, -2). D is a point on AC.

(a) Write down the coordinates of A and D.

(b) Find the coordinates of the centroid of ∆ABC.

(c) If D divides AC in the ratio k : 1, find the value of k.

(d) Find the equation of the line BD.

In the given diagram, ABC is a triangle, where B(4, -4) and C(-4, -2). D is a point on AC. ICSE 2024 Maths Solved Question Paper.
Answer

(a) From graph,

Co-ordinates of A = (0, 6) and D = (-3, 0)

(b) By formula,

Co-ordinates of centroid = (x1+x2+x33,y1+y2+y33)\Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big)

=(0+4+(−4)3,6+(−4)+(−2)3)=(03,03)=(0,0).= \Big(\dfrac{0 + 4 + (-4)}{3}, \dfrac{6 + (-4) + (-2)}{3}\Big) \\[1em] = \Big(\dfrac{0}{3}, \dfrac{0}{3}\Big) \\[1em] = (0, 0).

Hence, co-ordinates of centroid of ∆ABC = (0, 0).

(c) 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)

Given,

D divides AC in the ratio k : 1.

∴(−3,0)=(k×−4+1×0k+1,k×−2+1×6k+1)⇒(−3,0)=(−4k+0k+1,−2k+6k+1)⇒(−3,0)=(−4kk+1,−2k+6k+1)⇒−3=−4kk+1 and 0=−2k+6k+1⇒−3(k+1)=−4k and 0=−2k+6⇒3(k+1)=4k and 2k=6⇒3k+3=4k and k=62⇒4k−3k=3 and k=3⇒k=3.\therefore (-3, 0) = \Big(\dfrac{k \times -4 + 1 \times 0}{k + 1}, \dfrac{k \times -2 + 1 \times 6}{k + 1}\Big) \\[1em] \Rightarrow (-3, 0) = \Big(\dfrac{-4k + 0}{k + 1}, \dfrac{-2k + 6}{k + 1}\Big) \\[1em] \Rightarrow (-3, 0) = \Big(\dfrac{-4k}{k + 1}, \dfrac{-2k + 6}{k + 1}\Big) \\[1em] \Rightarrow -3 = -\dfrac{4k}{k + 1} \text{ and } 0 = \dfrac{-2k + 6}{k + 1} \\[1em] \Rightarrow -3(k + 1) = -4k \text{ and } 0 = -2k + 6 \\[1em] \Rightarrow 3(k + 1) = 4k \text{ and } 2k = 6 \\[1em] \Rightarrow 3k + 3 = 4k \text{ and } k = \dfrac{6}{2} \\[1em] \Rightarrow 4k - 3k = 3 \text{ and } k = 3 \\[1em] \Rightarrow k = 3.

Hence, k = 3.

(d) By two point form,

Equation of line :

y - y1 = y2−y1x2−x1(x−x1)\dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1)

Equation of BD :

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

⇒ y + 4 = 4−7(x−4)\dfrac{4}{-7}(x - 4)

⇒ -7(y + 4) = 4(x - 4)

⇒ -7y - 28 = 4x - 16

⇒ 4x + 7y - 16 + 28 = 0

⇒ 4x + 7y + 12 = 0.

Hence, equation of BD is 4x + 7y + 12 = 0.