Jamia Millia Islamia MCA Rectangular Cartesian Coordinates Previous Year Questions (PYQs)
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The locus of a point for which $y=0,\ z=0$ is:
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Solution
Median of equilateral triangle = $\dfrac{\sqrt{3}}{2} \times \text{side}$
$\Rightarrow 3a = \dfrac{\sqrt{3}}{2} \times \text{side}$
$\Rightarrow \text{side} = 2\sqrt{3}a$
Radius (distance from centre to vertex) = $\dfrac{\text{side}}{\sqrt{3}} = 2a$.
Equation: $x^2 + y^2 = (2a)^2 = 4a^2$.
$\boxed{\text{Answer: (C) }x^2 + y^2 = 4a^2}$
The reflection of the point $(\alpha, \beta, \gamma)$ in the xy-plane is:
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Solution
Reflection in xy-plane → z-coordinate changes sign, x and y remain same.
$\boxed{\text{Answer: (D) }(\alpha, \beta, -\gamma)}$
The locus represented by $xy + yz = 0$ is:
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Solution
Given equation: $xy + yz = 0 \Rightarrow y(x + z) = 0$
This represents two planes:
1️⃣ $y = 0$
2️⃣ $x + z = 0$
Normal to first plane = $(0,1,0)$
Normal to second plane = $(1,0,1)$
Their dot product = $0(1) + 1(0) + 0(1) = 0$,
so the planes are perpendicular.
$\boxed{\text{Answer: (D) A pair of perpendicular planes}}$
The distance of the point $(2, 5, 7)$ from the x-axis is……
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Solution
Distance from x-axis $= \sqrt{y^2 + z^2} = \sqrt{5^2 + 7^2} = \sqrt{74}$.
The distance between $P(x_1,y_1)$ and $Q(x_2,y_2)$ is given by $|x_2-x_1|$ when $PQ$ is …
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Solution
Distance equals $|x_2-x_1|$ when the segment is horizontal (same $y$), i.e., parallel to the $x$-axis.
Region represented by $x \ge 0,\ y \ge 0$ is
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Solution
Using triangle inequality,
Hence, value is less than 12.
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