(i) Distance from Heera Bagh C(4, 3) to High Court A(−4, 2):
$$AC = \sqrt{(4-(-4))^2 + (3-2)^2} = \sqrt{64+1} = \sqrt{65} \text{ units}$$
Daily distance (to court and back) = $2\sqrt{65}$ units
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(ii) Let the point on X-axis dividing AD in ratio k : 1 be (x, 0).
A = (−4, 2), D = (−5, −2)
Using section formula for y-coordinate:
$$0 = \frac{k(-2) + 1(2)}{k+1} \Rightarrow -2k + 2 = 0 \Rightarrow k = 1$$
∴ AD is divided in ratio 1 : 1
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(iii) B = (4, −4), C = (4, 3), D = (−5, −2)
$$BC = \sqrt{(4-4)^2+(-4-3)^2} = \sqrt{0+49} = 7 \text{ units}$$
$$BD = \sqrt{(4-(-5))^2+(-4-(-2))^2} = \sqrt{81+4} = \sqrt{85} \text{ units}$$
Since BC ≠ BD, Birla Mandir is NOT equidistant from Heera Bagh and Amar Jawan Jyoti.
Source: Coordinate Geometry, Distance Formula & Section Formula
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