NIMCET Rectangular Cartesian Coordinates Previous Year Questions (PYQs)
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If the distance of $(x,y)$ from the origin is defined as
$d(x,y) = \max(|x|,|y|)$,
then the locus of points where $d(x,y)=1$ is:
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Solution
$\max(|x|,|y|)=1$ describes a square with vertices $(\pm1,\pm1)$.
Side length = $2$
Area = $4$
$ABC$ is isosceles with $AB = AC$.
$BC$ is parallel to x-axis.
$m_1, m_2$ are slopes of the medians from $B$ and $C$.
Then:
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4
Solution
Let coordinates:
$B(-a, 0)$
$C(a, 0)$
$A(0, h)$
Median from $B$ goes to midpoint of $AC$: $(0, h/2)$
Slope: $m_1 = \dfrac{h/2 - 0}{0 - (-a)} = \dfrac{h}{2a}$
Median from $C$ goes to midpoint of $AB$: $(0, h/2)$
Slope: $m_2 = \dfrac{h/2 - 0}{0 - a} = -\dfrac{h}{2a}$
Thus:
$m_1 + m_2 = 0$
A line $L$ has intercepts $a$ and $b$ on the coordinate axes. When the axes are rotated through a given angle, keeping the origin fixed, the same line has intercepts $p$ and $q$. Which of the following is true?
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Solution
Intercept form of line remains invariant under rotation in terms of reciprocal squares.
Answer: $\boxed{\dfrac{1}{a^2}+\dfrac{1}{b^2}=\dfrac{1}{p^2}+\dfrac{1}{q^2}}$
The point $(4,1)$ undergoes the following transformations successively:
(i) Reflection about the line $y=x$
(ii) Translation through a distance $2$ units along the positive $x$-axis
(iii) Rotation by an angle $\frac{\pi}{4}$ anticlockwise about the origin
The final position of the point is:
1
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4
Solution
Solution:
Step 1: Reflect (4,1) about y=x → (1,4)
Step 2: Translate 2 units in +x direction → (1+2, 4) = (3,4)
Step 3: Rotate (3,4) by π/4 anticlockwise:
New x = (3 - 4)/√2 = -1/√2
New y = (3 + 4)/√2 = 7/√2
Final point = $\left(\frac{-1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$
If P(1,2), Q(4,6), R(5,7) and S(a,b) are the vertices of a parallelogram PQRS, then
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Solution
If the points 
lie in the region corresponding to the acute angle between the lines and then
lie in the region corresponding to the acute angle between the lines and then1
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Solution
If (2, 1), (–1, –2), (3, 3) are the midpoints of the sides BC, CA, AB of a triangle ABC, then
equation of the line BC is
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Solution
The median AD of ΔABC is bisected at E and BE is produced to meet the side AC at F. Then, AF ∶ FC is
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Solution
A triangle has a vertex at (1, 2) and the mid points of the two sides through it are (-1, 1) and (2, 3). Then the area of this triangle is
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Solution
Let vertex be $A(1,2)$.
Midpoints of sides through $A$ are:
$M_1(-1,1)$ and $M_2(2,3)$
Let other two vertices be $B(x_1,y_1)$ and $C(x_2,y_2)$.
Using midpoint formula:
$\left(\frac{1+x_1}{2},\frac{2+y_1}{2}\right)=(-1,1)$
$x_1=-3,\ y_1=0$
So, $B(-3,0)$
Similarly,
$\left(\frac{1+x_2}{2},\frac{2+y_2}{2}\right)=(2,3)$
$x_2=3,\ y_2=4$
So, $C(3,4)$
Area of triangle:
$\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$
$=\frac{1}{2}|1(0-4)+(-3)(4-2)+3(2-0)|$
$=\frac{1}{2}|-4-6+6|$
$=\frac{1}{2}\times 4$
$=2$
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