CUET PG MCA Rectangular Cartesian Coordinates Previous Year Questions (PYQs)
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Showing 5 questions
| List I | List II | ||
|---|---|---|---|
| A. | Kailash Satyarthi | I. | Chemistry |
| B. | Abhijit Banerjee | II. | Peace |
| C. | Vinkatraman Ramakrishnan | III. | Physics |
| D. | Subrahmanyan Chandrasekhar | IV. | Economics |
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Solution
| List I | List II | ||
|---|---|---|---|
| A. | Kailash Satyarthi | II. | Peace |
| B. | Abhijit Banerjee | IV. | Economics |
| C. | Venkatraman Ramakrishnan | I. | Chemistry |
| D. | Subrahmanyan Chandrasekhar | III. | Physics |
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: If $a \ne b$ then $(ab) \ne (b,a)$.
Reason R: $(4,-3)$ lies in quadrant IV.
In the light of the above statements, choose the correct answer from the options given below:
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Solution
Assertion A is true because ordered pairs depend on order.
Reason R is also true since $(+,-)$ lies in quadrant IV.
But R does not explain A.
If the vertices of a triangle are O(0,0), A(a,0) and B(0,a). Then, the distance between its circumcenter and orthocenter is:
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Solution
The points $(K,2-2K)$, $(-K+1,2K)$ and $(-4-K,6-2K)$ are collinear if:
(A) $K=\frac{1}{2}$
(B) $K=\frac{-1}{2}$
(C) $K=\frac{3}{2}$
(D) $K=-1$
(E) $K=1$
Choose the correct answer from the options given below:
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Solution
Given points:
\[
P_1(K,2-2K), \quad P_2(1-K,2K), \quad P_3(-4-K,6-2K)
\]
These three points are collinear if the area of triangle formed by them is zero:
\[
\Delta =
\begin{vmatrix}
K & 2-2K & 1\\
1-K & 2K & 1\\
-4-K & 6-2K & 1
\end{vmatrix} = 0
\]
Expanding determinant:
\[
8K^2 + 4K - 4 = 0 \;\;\Rightarrow\;\; 2K^2 + K - 1 = 0
\]
Solving quadratic:
\[
K = \frac{-1 \pm 3}{4} \;\;\Rightarrow\;\; K=\tfrac{1}{2},\; -1
\]
If the vertices of a triangle are (1, 2), (2, 5) and (4, 3) then the area of the triangle is:
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Solution
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