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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?
(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.
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?
(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.
Show SolutionHide Solution↓
Sol. (i) Distance travelled $= 2 AC$
$= 2\sqrt{(-4-4)^2 + (2 - 3)^2}$ (I) ($\frac{1}{2}$ Mark)
$= 2\sqrt{64 + 1}$ (II) ($\frac{1}{2}$ Mark)
$= 2\sqrt{65}$
Hence, required distance is $2\sqrt{65}$ units.
(ii) Let the point $P(x, 0)$ divides $AD$ in the ratio $K : 1$
$\therefore AP: PD = K : 1$
Here, $0 = \frac{-2K+2}{K+1}$ (I) ($\frac{1}{2}$ Mark)
$\Rightarrow K = 1$ (II) ($\frac{1}{2}$ Mark)
$\therefore$ The required ratio is $1 : 1$
(iii) (a) $BC = \sqrt{(4 - 4)^2 + (-4 - 3)^2} = 7$ units (I) (1 Mark)
$BD = \sqrt{(4 + 5)^2 + (-4 + 2)^2} = \sqrt{85}$ units (II) (1 Mark)
$\therefore BC \neq BD$
$\therefore$ Birla Mandir is not equidistant from Heera Bagh and Amar Jawan Jyoti.
OR
(b) Let us assume that points $A, O, B$ are collinear and $AO : OB = K : 1$
Here, $0 = \frac{4K-4}{K+1}$ (I) ($\frac{1}{2}$ Mark)
$\Rightarrow K = 1$ (II) ($\frac{1}{2}$ Mark)
Also, $0 = \frac{-4K+2}{K+1}$ (III) ($\frac{1}{2}$ Mark)
$\Rightarrow K = \frac{1}{2}$ (IV) ($\frac{1}{2}$ Mark)
Since the value of $K$ is different in the above two cases, so points $A, O$ and $B$ are not collinear.
$= 2\sqrt{(-4-4)^2 + (2 - 3)^2}$ (I) ($\frac{1}{2}$ Mark)
$= 2\sqrt{64 + 1}$ (II) ($\frac{1}{2}$ Mark)
$= 2\sqrt{65}$
Hence, required distance is $2\sqrt{65}$ units.
(ii) Let the point $P(x, 0)$ divides $AD$ in the ratio $K : 1$
$\therefore AP: PD = K : 1$
Here, $0 = \frac{-2K+2}{K+1}$ (I) ($\frac{1}{2}$ Mark)
$\Rightarrow K = 1$ (II) ($\frac{1}{2}$ Mark)
$\therefore$ The required ratio is $1 : 1$
(iii) (a) $BC = \sqrt{(4 - 4)^2 + (-4 - 3)^2} = 7$ units (I) (1 Mark)
$BD = \sqrt{(4 + 5)^2 + (-4 + 2)^2} = \sqrt{85}$ units (II) (1 Mark)
$\therefore BC \neq BD$
$\therefore$ Birla Mandir is not equidistant from Heera Bagh and Amar Jawan Jyoti.
OR
(b) Let us assume that points $A, O, B$ are collinear and $AO : OB = K : 1$
Here, $0 = \frac{4K-4}{K+1}$ (I) ($\frac{1}{2}$ Mark)
$\Rightarrow K = 1$ (II) ($\frac{1}{2}$ Mark)
Also, $0 = \frac{-4K+2}{K+1}$ (III) ($\frac{1}{2}$ Mark)
$\Rightarrow K = \frac{1}{2}$ (IV) ($\frac{1}{2}$ Mark)
Since the value of $K$ is different in the above two cases, so points $A, O$ and $B$ are not collinear.