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ICSE Class 10 Mathematics Question 13 of 21

Solved 2023 Question Paper ICSE Class 10 Mathematics — Question 9

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Question

Question 6(iii)

A and B are two points on the x-axis and y-axis respectively.

A and B are two points on the x-axis and y-axis respectively. ICSE 2023 Maths Solved Question Paper.

(a) Write down the co-ordinates of A and B.

(b) P is a point on AB such that AP : PB = 3 : 1. Using section formula find the coordinates of point P.

(c) Find the equation of a line passing through P and perpendicular to AB.

Answer

(a) From figure,

A = (4, 0) and B = (0, 4).

(b) Let coordinates of P 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)=(3×0+1×43+1,3×4+1×03+1)=(0+44,12+04)=(44,124)=(1,3).\Rightarrow (x, y) = \Big(\dfrac{3 \times 0 + 1 \times 4}{3 + 1}, \dfrac{3 \times 4 + 1 \times 0}{3 + 1}\Big) \\[1em] = \Big(\dfrac{0 + 4}{4}, \dfrac{12 + 0}{4}\Big) \\[1em] = \Big(\dfrac{4}{4}, \dfrac{12}{4}\Big) \\[1em] = (1, 3).

Hence, coordinates of P = (1, 3).

(c) By formula,

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

Substituting values we get :

Slope of AB = 4−00−4=4−4\dfrac{4 - 0}{0 - 4} = \dfrac{4}{-4} = -1.

We know that,

Product of slope of perpendicular lines = -1.

∴ Slope of AB × Slope of line perpendicular to AB = -1

⇒ -1 × Slope of line perpendicular to AB = -1

⇒ Slope of line perpendicular to AB = −1−1\dfrac{-1}{-1} = 1.

Line passing through P and perpendicular to AB :

⇒ y - y1 = m(x - x1)

⇒ y - 3 = 1(x - 1)

⇒ y - 3 = x - 1

⇒ y = x - 1 + 3

⇒ y = x + 2.

Hence, required equation is y = x + 2.