Question

Observe the map of Jaipur city placed on a Cartesian plane. Taking Rambagh Palace as origin, the location of some places are given below: Point A: (–4, 2) Rajasthan High Court Point B: (4, –4) Birla Mandir  Point C: (4, 3) Heera Bagh Point D: (–5, –2) Amar Jawan Jyoti

Based on the above, answer the following questions:

(i) Advocate Rehana stays at Heera Bagh. How much distance she has to cover daily to go to the court and coming back home?   [1]

(ii) There is a crossing on X-axis which divides AD in a certain ratio. Find the ratio.   [1]

(iii) (a) Is Birla Mandir equidistant from Heera Bagh and Amar Jawan Jyoti? Justify your answer.   [2]

OR

(b) Using section formula, show that points A, O and B are not collinear.

Answer 1

(i) Given:

A = (–4, 2)

C = (4, 3)

One-way distance (CA):

Distance `d_(CA) = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2` 

`d_(CA) = sqrt((-4 - 4)^2 + (2 - 3)^2`

`d_(CA) = sqrt((-8)^2 + (-1)^2`

`d_(CA) = sqrt(64 + 1)`

`d_(CA) = sqrt(65)` units

Daily total distance (Round trip) = `2 xx sqrt(65) = 2sqrt(65)` units.

The total distance covered dally is `2sqrt(65)` units.

(ii) Let the ratio be k : 1. The point on the X-axis is (x, 0).

Point A is (4, 2) and Point D is (–5, –2).

Using the Y-coordinate:

`y = (my_2 + ny_1)/(m + n)`

`0 = (k(-2) + 1(2))/(k + 1)`

0 = –2k + 2

2k = 2

k = 1

Thus, the ratio is 1 : 1.

(iii) (a) Equidistant means the distance from Birla Mandir (B) to Heera Bagh (C) must be equal to the distance from Birla Mandir (B) to Amar Jawan Jyoti (D).

B = (4, –4), C = (4, 3) and D = (–5, –2)

Distance BC:

`BC = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`

`BC = sqrt((4 - 4)^2 + (3 - (-4))^2`

`BC = sqrt(0^2 + 7^2)`

BC = 7 units

Distance BD:

`BD = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`

`BD = sqrt((-5 - 4)^2 + (-2 - (-4))^2`

`BD = sqrt((-9)^2 + 2^2)`

`BD = sqrt(81 + 4)`

`BD = sqrt(85)` units

Since `7 ≠ sqrt(85)`, BC ≠ BD.

No, Birla Mandir is not equidistant from the two places because the distances calculated are unequal.

OR

(b) Assume O divides AB in ratio k : 1.

A = (–4, 2), O = (0, 0) and B = (4, –4)

Using the X-coordinate of O:

`0 = (k(4) + 1(-4))/(k + 1)`

4k – 4 = 0

⇒ k = 1

If collinear, O must divide AB in ratio 1 : 1. 

Let’s check this ratio using the Y-coordinate. If k = 1, the Y-coordinate should be:

`y = (1(-4) + 1(2))/(1 + 1)`

= `(-2)/2`

= –1

But the Y-coordinate of O is 0. Since 0 ≠ –1, the point O does not lie on the line segment AB in a single consistent ratio.

Since the ratios for X and Y coordinates do not match, the points A, O and B are not collinear.