ICSE Class 10 Maths Chapter 13 Distance and Section Formulae

Chapter 13

Distance and Section Formulae

Points To Remember

1. Distance Formula

Theorem 1. Show that the distance between the points P \((x_1, y_1)\) and Q \((x_2, y_2)\) is given by the formula:

\[PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]

Proof. Let X'OX and YOY' be the co-ordinate axes. Let P \((x_1, y_1)\) and Q \((x_2, y_2)\) be the given points in the plane.

Draw PM and QN perpendiculars on x-axis.

Also, draw PR \(\perp\) NQ. Then,

OM = \(x_1\), ON = \(x_2\), PM = \(y_1\) and QN = \(y_2\).

\(\therefore\) PR = MN = ON - OM = \((x_2 - x_1)\).

QR = (QN - RN) = \((y_2 - y_1)\). [- RN = PM = \(y_1\)]

Now, from right-angled \(\triangle\)PQR, by Pythagoras Theorem, we have

PQ\(^2\) = PR\(^2\) + QR\(^2\)

= \((x_2 - x_1)^2 + (y_2 - y_1)^2\).

\(\therefore\) \(PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Corollary: The distance of a point P \((x, y)\) from the origin O \((0, 0)\) is given by:

\[OP = \sqrt{(x-0)^2 + (y-0)^2} = \sqrt{x^2 + y^2}\]

2. Section Formula

Theorem 2: Prove that the co-ordinates of the points P \((x, y)\) which divides the line joining A \((x_1, y_1)\) and B \((x_2, y_2)\) internally in the ratio m : n are given by

\[x = \frac{mx_2 + nx_1}{m+n}, \quad y = \frac{my_2 + ny_1}{m+n}\]

Proof: Let A \((x_1, y_1)\) and B \((x_2, y_2)\) be the given points and let P \((x, y)\) be the point which divides AB in the ratio m : n.

Then, \(\frac{AP}{PB} = \frac{m}{n}\).

Draw AL, BM and PQ perpendiculars on x-axis.

Also, draw AR \(\perp\) PQ and PS \(\perp\) BM. Then,

AR = LQ = OQ - OL = \((x - x_1)\).

PS = QM = OM - OQ = \((x_2 - x)\).

PR = PQ - RQ = PQ - AL = \((y - y_1)\).

BS = BM - SM = BM - PQ = \((y_2 - y)\).

Clearly, \(\triangle\)ARP and \(\triangle\)PSB are similar and therefore, their sides are proportional.

\(\therefore\) \(\frac{AP}{PB} = \frac{AR}{PS} = \frac{PR}{BS}\)

Now, \(\frac{AP}{PB} = \frac{AR}{PS} \Rightarrow \frac{m}{n} = \frac{x - x_1}{x_2 - x} \Rightarrow m(x_2 - x) = n(x - x_1) \Rightarrow x = \left(\frac{mx_2 + nx_1}{m+n}\right)\).

Again, \(\frac{AP}{PB} = \frac{PR}{BS} \Rightarrow \frac{m}{n} = \frac{y - y_1}{y_2 - y} \Rightarrow m(y_2 - y) = n(y - y_1)\)

\(\Rightarrow\) \(y = \left(\frac{my_2 + ny_1}{m+n}\right)\).

Hence, the co-ordinates of P are \(\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)\).

Corollary: Show that the co-ordinates of the mid-point M of a line segment with end points A \((x_1, y_1)\) and B \((x_2, y_2)\) are: \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\).

Proof. Let M be the mid-point of the line segment joining the points A \((x_1, y_1)\) and B \((x_2, y_2)\). Then, M divides AB in the ratio 1 : 1.

\(\therefore\) Co-ordinates of M are \(\left(\frac{1 \cdot x_2 + 1 \cdot x_1}{1+1}, \frac{1 \cdot y_2 + 1 \cdot y_1}{1+1}\right)\), i.e., \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\).

Hence, the co-ordinates of the mid-point of AB are \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\).

3. Centroid of a Triangle: The point of intersection of the medians of a triangle is called its centroid.

To Find the Co-ordinates of the Centroid of a Triangle.

Let A\((x_1, y_1)\), B\((x_2, y_2)\) and C\((x_3, y_3)\) be the vertices of a given \(\triangle\)ABC. Let D be the mid-point of BC.

Then, the co-ordinates of D are \(\left(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}\right)\)

Let G \((x, y)\) be the centroid of \(\triangle\)ABC. Then, G divides AD in the ratio 2 : 1.

\(\therefore\) \(x = \frac{2 \cdot \frac{(x_2 + x_3)}{2} + 1 \cdot x_1}{2+1} = \left(\frac{x_1 + x_2 + x_3}{3}\right)\).

\(\text{and } y = \frac{2 \cdot \frac{(y_2 + y_3)}{2} + 1 \cdot y_1}{2+1} = \left(\frac{y_1 + y_2 + y_3}{3}\right)\)

Hence, the co-ordinates of G are \(\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)\) \(\left(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}\right)\)

Exercise 13 (A)

Q.1. Find the distance between each of the following pairs of points:

(i) A \((8, 3)\) and B \((14,11)\)

(ii) A \((3, -5)\) and B \((8, 7)\)

(iii) P \((2, -3)\) and Q \((-6, 3)\)

(iv) P \((-6, -4)\) and Q \((9, 4)\)

(v) M \((-8, -3)\) and N \((- 2, -5)\)

(vi) R \((a + b, a - b)\) and S \((a - b, a + b)\),

Sol. We know that, distance between two points A \((x_1, y_1)\) and B \((x_2, y_2)\)

\(= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

(i) \(\therefore\) Distance between A \((8, 3)\) and B \((14, 11)\)

\(= \sqrt{(14-8)^2 + (11-3)^2}\)

\(= \sqrt{(6)^2 + (8)^2} = \sqrt{36+64}\)

\(= \sqrt{100} = 10 \text{ units } \mathbf{Ans.}\)

(ii) Distance between A \((3, -5)\) and B \((8, 7)\)

\(= \sqrt{(8-3)^2 + [7-(-5)]^2}\)

\(= \sqrt{(8-3)^2 + (7+5)^2}\)

\(= \sqrt{(5)^2 + (12)^2} = \sqrt{25+144}\)

\(= \sqrt{169} =13 \text{ units. } \mathbf{Ans.}\)

(iii) Distance between P \((2, -3)\) and Q \((-6, 3)\)

\(= \sqrt{(-6,-2)^2 + [3-(-3)]^2}\)

\(= \sqrt{(-8)^2 + (3+3)^2} = \sqrt{(-8)^2 + (6)^2}\)

\(= \sqrt{64+36} = \sqrt{100} = 10 \text{ units } \mathbf{Ans.}\)

(iv) Distance between P \((-6, -4)\) and Q \((9, 4)\)

\(= \sqrt{[9-(-6)]^2 + [4-(-4)]^2}\)

\(= \sqrt{(9+6)^2 + (4+4)^2} = \sqrt{(15)^2 + (8)^2}\)

\(= \sqrt{225+64} = \sqrt{289} =17 \text{ units } \mathbf{Ans.}\)

(v) Distance between M \((-8, -3)\) and N \((-2, -5)\)

\(= \sqrt{[-2-(-8)]^2 + [-5-(-3)]^2}\)

\(= \sqrt{(-2+8)^2 + (-5+3)^2}\)

\(= \sqrt{(6)^2 + (-2)^2} = \sqrt{36+4} = \sqrt{40} = \sqrt{4 \times 10}\)

\(= 2\sqrt{10} \text{ units } \mathbf{Ans.}\)

(vi) Dis tance between R \((a + b, a - b)\) and S \((a - b, a + b)\).

\(= \sqrt{(a-b-a-b)^2 + (a+b-a+b)^2}\)

\(= \sqrt{(-2b)^2 + (2b)^2} = \sqrt{4b^2 + 4b^2} = \sqrt{8b^2}\)

\(= \sqrt{4b^2 \times 2} = 2b\sqrt{2} = 2\sqrt{2b} \text{ units } \mathbf{Ans.}\)

Teacher's Note

Distance formula helps us find how far apart two locations are on a map, like calculating the straight-line distance between two cities using their coordinates on a grid.

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