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NIMCET Previous Year Questions (PYQs)

NIMCET Rectangular Cartesian Coordinates PYQ

NIMCET PYQ
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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NIMCET Previous Year PYQ NIMCET NIMCET 2009 PYQ

Solution

$\max(|x|,|y|)=1$ describes a square with vertices $(\pm1,\pm1)$. 
Side length = $2$ 
Area = $4$
NIMCET PYQ
$ABC$ 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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NIMCET Previous Year PYQ NIMCET NIMCET 2009 PYQ

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: $m_1 + m_2 = 0$
NIMCET PYQ
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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NIMCET Previous Year PYQ NIMCET NIMCET 2008 PYQ

Solution

Intercept form of line remains invariant under rotation in terms of reciprocal squares. Answer: $\boxed{\dfrac{1}{a^2}+\dfrac{1}{b^2}=\dfrac{1}{p^2}+\dfrac{1}{q^2}}$
NIMCET PYQ
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:





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NIMCET Previous Year PYQ NIMCET NIMCET 2010 PYQ

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 = $\left(\frac{-1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$
NIMCET PYQ
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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NIMCET Previous Year PYQ NIMCET NIMCET 2021 PYQ

Solution

NIMCET PYQ
If the points  lie in the region corresponding to the acute angle between the lines and then





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NIMCET Previous Year PYQ NIMCET NIMCET 2019 PYQ

Solution


NIMCET PYQ
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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NIMCET Previous Year PYQ NIMCET NIMCET 2015 PYQ

Solution

NIMCET PYQ
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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NIMCET Previous Year PYQ NIMCET NIMCET 2014 PYQ

Solution


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