Comprehension
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:
Question: 1

Advocate Rehana stays at Heera Bagh. How much distance she has to cover daily to go to the court and coming back home?

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Don't forget to multiply by 2 for round-trip questions! Read the wording carefully for "daily" or "one-way".
Updated On: Feb 20, 2026
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Solution and Explanation

Step 1: Understanding the Concept:
The distance between two points in a Cartesian plane is found using the distance formula. The total daily distance is twice the distance between the home and the court.
Step 2: Key Formula or Approach:
Distance \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).
Home (C) is \((4, 3)\) and Court (A) is \((-4, 2)\).
Step 3: Detailed Explanation:
One-way distance (CA):
\[ d_{CA} = \sqrt{(-4 - 4)^2 + (2 - 3)^2} \]
\[ d_{CA} = \sqrt{(-8)^2 + (-1)^2} \]
\[ d_{CA} = \sqrt{64 + 1} = \sqrt{65} \text{ units} \]
Daily total distance (Round trip) = \(2 \times \sqrt{65} = 2\sqrt{65}\) units.
Step 4: Final Answer:
The total distance covered daily is \(2\sqrt{65}\) units.
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Question: 2

There is a crossing on X-axis which divides AD in a certain ratio. Find the ratio.

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Points on the X-axis have \(y = 0\); points on the Y-axis have \(x = 0\). Use this property to eliminate one variable when finding ratios.
Updated On: Feb 20, 2026
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Solution and Explanation

Step 1: Understanding the Concept:
A point on the X-axis always has a Y-coordinate of 0. We can use the section formula to find the ratio.
Step 2: Key Formula or Approach:
Section formula for Y-coordinate: \(y = \frac{m y_2 + n y_1}{m + n}\).
Point A is \((-4, 2)\) and Point D is \((-5, -2)\).
Step 3: Detailed Explanation:
Let the ratio be \(k : 1\). The point on the X-axis is \((x, 0)\).
Using the Y-coordinate:
\[ 0 = \frac{k(-2) + 1(2)}{k + 1} \]
\[ 0 = -2k + 2 \]
\[ 2k = 2 \]
\[ k = 1 \]
Thus, the ratio is \(1 : 1\).
Step 4: Final Answer:
The ratio is \(1 : 1\).
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Question: 3

Is Birla Mandir equidistant from Heera Bagh and Amar Jawan Jyoti? Justify your answer.

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Square the distances to compare them faster: \(7^2 = 49\) while \((\sqrt{85})^2 = 85\). This avoids calculating square roots of non-perfect squares.
Updated On: Feb 20, 2026
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Solution and Explanation

Step 1: Understanding the Concept:
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).
Step 2: Key Formula or Approach:
Calculate distances \(BC\) and \(BD\) using the distance formula: \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).
B: \((4, -4)\), C: \((4, 3)\), D: \((-5, -2)\).
Step 3: Detailed Explanation:
Calculating distance BC:
\[ BC = \sqrt{(4 - 4)^2 + (3 - (-4))^2} \]
\[ BC = \sqrt{0^2 + 7^2} = 7 \text{ units} \]
Calculating distance BD:
\[ BD = \sqrt{(-5 - 4)^2 + (-2 - (-4))^2} \]
\[ BD = \sqrt{(-9)^2 + 2^2} \]
\[ BD = \sqrt{81 + 4} = \sqrt{85} \text{ units} \]
Since \(7 \neq \sqrt{85}\), \(BC \neq BD\).
Step 4: Final Answer:
No, Birla Mandir is not equidistant from the two places because the distances calculated are unequal.
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Question: 4

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

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Collinearity can also be checked using slopes. Slope \(AO = \frac{0-2}{0-(-4)} = -1/2\). Slope \(OB = \frac{-4-0}{4-0} = -1\). Since slopes are different, points are not collinear.
Updated On: Feb 20, 2026
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Solution and Explanation

Step 1: Understanding the Concept:
Points are collinear if one point divides the segment joining the other two in some ratio. If the ratios calculated for X and Y coordinates differ, the points are not collinear.
Step 2: Key Formula or Approach:
A: \((-4, 2)\), O: \((0, 0)\), B: \((4, -4)\).
Assume O divides AB in ratio \(k:1\).
Step 3: Detailed Explanation:
Using the X-coordinate of O:
\[ 0 = \frac{k(4) + 1(-4)}{k + 1} \]
\[ 4k - 4 = 0 \implies k = 1 \]
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:
\[ y = \frac{1(-4) + 1(2)}{1 + 1} = \frac{-2}{2} = -1 \]
But the Y-coordinate of O is 0. Since \(0 \neq -1\), the point O does not lie on the line segment AB in a single consistent ratio.
Step 4: Final Answer:
Since the ratios for X and Y coordinates do not match, the points A, O, and B are not collinear.
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