Distance Formula — Case-Study Based Question

A group of class IX students goes to a picnic during winter holidays. The positions of three friends Nitin, Rajesh and Kareem are shown by the points P, Q and R respectively.

(i) Find the distance between :

(a) Nitin and Rajesh

(b) Rajesh and Kareem

(c) Nitin and Kareem

(ii) Show that P, Q and R are collinear.

(iii) Find the point on x-axis which is equidistant from points Q and R.

(iv) If R is taken as origin; what will be the co-ordinates of P and Q ?

Answer

Given,

The co-ordinates are :

P(Nitin): (6, 4)

Q(Rajesh): (11, 9)

R(Kareem): (9, 7)

We know that

Distance between the given points = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

(i) (a) Distance between Nitin(P) and Rajesh(Q),

$PQ = \sqrt{(11 - 6)^2 + (9 - 4)^2} \\[1em] = \sqrt{5^2 + 5^2} \\[1em] = \sqrt{25 + 25} \\[1em] = \sqrt{50} \\[1em] = 5\sqrt{2}.$

Hence, distance between Nitin and Rajesh = $5\sqrt{2}$ units.

(b) Distance between Rajesh(Q) and Kareem(R),

$QR = \sqrt{(11 - 9)^2 + (9 - 7)^2} \\[1em] = \sqrt{2^2 + 2^2} \\[1em] = \sqrt{8} \\[1em] = 2\sqrt{2}.$

Hence, distance between Rajesh and Kareem = $2\sqrt{2}$ units.

(c) Distance between Nitin(P) and Kareem(R),

$PR = \sqrt{(9 - 6)^2 + (7 - 4)^2} \\[1em] = \sqrt{3^2 + 3^2} \\[1em] = \sqrt{18} \\[1em] = 3\sqrt{2}.$

Hence, distance between Nitin and Kareem = $3\sqrt{2}$ units.

(ii) Points P, Q and R are collinear only if, PR + RQ = PQ

Consider L.H.S,

PR + RQ = $3\sqrt{2} + 2\sqrt{2}$

= $5\sqrt{2}$

Consider R.H.S,

PQ = $5\sqrt{2}$

∴ L.H.S = R.H.S

So, they are collinear.

Hence, P, Q and R are collinear.

(iii) Let S (x, 0) be the point on x-axis equidistant from Q and R.

According to question,

SQ = SR

$\sqrt{(x - 11)^2 + (0 - 9)^2} = \sqrt{(x - 9)^2 + (0 - 7)^2}$

Squaring on both sides,

(x - 11)² + (0 - 9)² = (x - 9)² + (0 - 7)²

x² - 22x + 121 + 81 = x² - 18x + 81 + 49

-18x + 22x = 121 + 81 - 81 - 49

4x = 121 - 49

4x = 72

x = $\dfrac{72}{4}$ = 18.

∴ The point S = (18, 0).

Hence, point on x-axis that is equidistant from Q and R = (18, 0).

(iv) If R is taken as origin, to find points of P and Q, their coordinates must be subtracted from the original coordinates of R(9, 7).

P' = (6 - 9, 4 - 7) = (-3, -3)

Q' = (11 - 9, 9 - 7) = (2, 2)

Hence, coordinates of P and Q are = (-3, -3) and (2, 2) respectively.

Rohan and Sohan stay in a village which are 50 km apart and are situated on Delhi Agra Highway as shown in the figure. Another highway YY' crosses Agra Delhi Highway at O (0, 0). A small local road PQ crosses both the highways at points A and B such that OA = 10 km and OB = 12 km. Also the villages of Sita and Gita are on the smaller highway YY'. Sita's village is 12 km from O and Gita's village is 15 km from O.

Now answer the following questions:

(i) Find the Coordinates of A and B

(ii) Find the lengths of local road AB.

(iii) What is the shortest distance Sohan will cover to reach Sita's village.

(iv) How far is Rohan's village from Gita's village ?

Answer

(i) Point A is on the X-axis at a distance of 10 km from O (0, 0). So, A = (10, 0).

Point B is on the Y-axis at a distance of 12 km from O (0, 0). So, B = (0, 12).

Hence, coordinates of A = (10, 0) and coordinates of B = (0, 12).

(ii) Using Pythagoras theorem for the △OAB:

AB² = BO² + OA²

AB² = 12² + 10²

AB² = 144 + 100

AB² = 244

AB = $\sqrt{244}$

AB = 15.6 km.

Hence, length of the local road AB = 15.6 km.

(iii) Based on the diagram, Sohan's village is at (30, 0) and Sita's village is at (0, 12).

$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] = \sqrt{(30 - 0)^2 + (0 - 12)^2} \\[1em] = \sqrt{30^2 + (-12)^2} \\[1em] = \sqrt{900 + 144} \\[1em] = \sqrt{1044} \\[1em] = 32.31 \text{ km}$

Hence, shortest distance between Sohan's village and Sita's village = 32.31 km.

(iv) Distance of Rohan's village from Gita's village:

Distance between Rohan's village to O + Distance between Gita's village to O

= 20 + 15

= 35 km.

Hence, distance from Rohan's village to Gita's village = 35 km.