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Mathematics · 2026 · 4 marks
CBSE 2026 · Region 4 · Set 2 · Q37
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:…
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 : $\displaystyle (-4,2)$ Rajasthan High Court Point B : $\displaystyle (4, -4)$ Birla Mandir Point C : $\displaystyle (4,3)$ Heera Bagh Point D : $\displaystyle (-5, -2)$ 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 ?
(ii)
There is a crossing on X-axis which divides AD in a certain ratio. Find the ratio.
(iii)
(a)
Is Birla Mandir equidistant from Heera Bagh and Amar Jawan Jyoti ? Justify your answer.
OR
(b)
Using section formula, show that points A, O and B are not collinear.
Marking-scheme solution
(i)
Distance travelled = $\displaystyle 2$ AC
\[\begin{array}{l}
=2 \sqrt{(-4-4)^{2}+(2-3)^{2}} \\
=2 \sqrt{64+1} \\
=2 \sqrt{65}
\end{array}
\]
Hence, required distance is $\displaystyle 2 \sqrt{65}$ units.
(ii)
Let the point $\displaystyle \mathrm{P}(x, 0)$ divides AD in the ratio K : $\displaystyle 1$
\[\therefore \mathrm{AP}: \mathrm{PD}=\mathrm{K}: 1
\]
Here, $\displaystyle 0=\frac{-2 \mathrm{~K}+2}{\mathrm{~K}+1}$
\[\Rightarrow \mathrm{K}=1
\]
∴ The required ratio is $\displaystyle 1$: $\displaystyle 1$
(iii)(a)
$\displaystyle \mathrm{BC}=\sqrt{(4-4)^{2}+(-4-3)^{2}}=7$ units
\[\begin{aligned}
& \mathrm{BD}=\sqrt{(4+5)^{2}+(-4+2)^{2}}=\sqrt{85} \text { units } \\
\therefore & \mathrm{BC} \neq \mathrm{BD}
\end{aligned}
\]
⇒ Birla Mandir is not equidistant from Heera Bagh and Amar Jawan Jyoti.
Let us assume that points $\displaystyle \mathrm{A}, \mathrm{O}, \mathrm{B}$ are collinear and $\displaystyle \mathrm{AO}: \mathrm{OB}=\mathrm{K}: 1$
\[\begin{array}{l}
\text { Here, } 0=\frac{4 \mathrm{~K}-4}{\mathrm{~K}+1} \\
\Rightarrow \mathrm{~K}=1 \\
\text { Also, } 0=\frac{-4 \mathrm{~K}+2}{\mathrm{~K}+1} \\
\Rightarrow \mathrm{~K}=\frac{1}{2}
\end{array}
\]
Since the value of K is different in the above two cases, so points A, O and B are not collinear.
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