NIMCET 2027 Straight line Previous Year Question Paper and Solution with PDF

Aspire's Library

A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations

NIMCET 2027
Previous Year Questions

NIMCET 2027
Previous Year Questions
NIMCET 2027

NIMCET 2027 Previous Year Questions (PYQs)

NIMCET 2027 Straight Line PYQ

Qus : 1 NIMCET 2027 PYQ

Let a, b, c, d be no zero numbers. If the point of intersection of the line 4ax + 2ay + c = 0 & 5bx + 2by + d=0 lies in the fourth quadrant and is equidistance from the two are then

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2023 PYQ

Solution

Given: Lines $4ax+2ay+c=0$ and $5bx+2by+d=0$.

Condition: Intersection lies in the 4th quadrant and is equidistant from the axes $\Rightarrow$ point is of the form $(t,-t)$ with $t>0$.

Substitute $y=-x$ in the two lines:
$4ax+2a(-x)+c=0 \;\Rightarrow\; 2x+\dfrac{c}{2a}=0 \;\Rightarrow\; x=-\dfrac{c}{2a}.$
$5bx+2b(-x)+d=0 \;\Rightarrow\; 3x+\dfrac{d}{b}=0 \;\Rightarrow\; x=-\dfrac{d}{3b}.$

Equate $x$ from both: $-\dfrac{c}{2a}=-\dfrac{d}{3b} \;\Rightarrow\; \boxed{3bc=2ad}.$

(QIV check: $x>0,\ y<0 \;\Rightarrow\; \dfrac{c}{a}<0$ and $\dfrac{d}{b}<0$ which is consistent.)

Qus : 2 NIMCET 2027 PYQ

The range of values of θ in the interval (0, π) such that the points (3,5) and (sinθ, cosθ) lie on the same side of the line x + y − 1 = 0, is

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2023 PYQ

Solution

Same Side of a Line — Geometric Condition

Line: $ x + y - 1 = 0 $
Point 1: (3, 2) → Lies on the side where value is positive:
\[ f(3, 2) = 3 + 2 - 1 = 4 > 0 \]

Point 2: $ (\cos\theta, \sin\theta) $ lies on same side if: \[ \cos\theta + \sin\theta > 1 \] Using identity: \[ \cos\theta + \sin\theta = \sqrt{2} \sin\left(\theta + \frac{\pi}{4}\right) \Rightarrow \sin\left(\theta + \frac{\pi}{4}\right) > \frac{1}{\sqrt{2}} \] So: \[ \theta \in \left(0,\ \frac{\pi}{2}\right) \]

✅ Final Answer: $ \boxed{\theta \in \left(0,\ \frac{\pi}{2}\right)} $

Qus : 3 NIMCET 2027 PYQ

Area enclosed by $|x|+|y|=1$

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2011 PYQ

Solution

This is a diamond (square rotated 45°) with diagonals length $2$ and $2$. Area $=\dfrac12 d_1 d_2=\dfrac12\cdot2\cdot2=2$

Qus : 4 NIMCET 2027 PYQ

The straight lines $\dfrac{x}{a} + \dfrac{y}{b} = k$ and $\dfrac{x}{a} + \dfrac{y}{b} = \dfrac{1}{k}$ (with $k\neq0$) meet on:

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2009 PYQ

Solution

Intersection point $(x,y)$ satisfies:

$\dfrac{x}{a} + \dfrac{y}{b} = k$

$\dfrac{x}{a} + \dfrac{y}{b} = \dfrac{1}{k}$

Subtract → inconsistent unless the meeting point satisfies: Multiply equations: $\left(\dfrac{x}{a} + \dfrac{y}{b}\right)\left(\dfrac{x}{a} + \dfrac{y}{b}\right) = 1$

Thus locus: $\left(\dfrac{x}{a} + \dfrac{y}{b}\right)^2 = 1$

This is a pair of parallel lines (degenerate hyperbola) → classified as hyperbola.

Qus : 5 NIMCET 2027 PYQ

Lines $2x + 3y - 6 = 0$ and $9x + 6y - 18 = 0$ cut coordinate axes in concyclic points. Center of circle is:

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2011 PYQ

Solution

Axis intercepts of both lines lie on same circle. Correct center is $ \left(\dfrac52,\dfrac52\right) $.

Qus : 6 NIMCET 2027 PYQ

In a parallelogram ABCD, P is the midpoint of AD. Also, BP and AC intersect at Q. Then AQ : QC =

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2019 PYQ

Solution

Qus : 7 NIMCET 2027 PYQ

The median AD of ΔABC is bisected at E and BE is extended to meet the side AC in F. The AF : FC =

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2019 PYQ

Solution

Qus : 8 NIMCET 2027 PYQ

If the straight line ax + by + c = 0 always passes through (1, –2), then a, b, c are in

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2013 PYQ

Solution

Qus : 9 NIMCET 2027 PYQ

The equations of the line parallel to the line $2x – 3y = 7$ and passing through the middle point of the line segment joining the points (1, 3) and (1, –7) is.

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2013 PYQ

Solution

Qus : 10 NIMCET 2027 PYQ

If C is the midpoint of AB and P is any point outside AB, then

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2016 PYQ

Solution

Qus : 11 NIMCET 2027 PYQ

If the perpendicular bisector of the line segment joining p(1,4) and q(k,3) has yintercept -4, then the possible values of k are

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2024 PYQ

Solution

Given: Points: $ P(1, 4) $, $ Q(k, 3) $

Step 1: Find midpoint of PQ

Midpoint = $ \left( \dfrac{1 + k}{2}, \dfrac{4 + 3}{2} \right) = \left( \dfrac{1 + k}{2}, \dfrac{7}{2} \right) $

Step 2: Find slope of PQ

Slope of PQ = $ \dfrac{3 - 4}{k - 1} = \dfrac{-1}{k - 1} $

Step 3: Slope of perpendicular bisector = negative reciprocal = $ k - 1 $

Step 4: Use point-slope form for perpendicular bisector:

$ y - \dfrac{7}{2} = (k - 1)\left(x - \dfrac{1 + k}{2}\right) $

Step 5: Find y-intercept (put $ x = 0 $)

$ y = \dfrac{7}{2} + (k - 1)\left( -\dfrac{1 + k}{2} \right) $

$ y = \dfrac{7}{2} - (k - 1)\left( \dfrac{1 + k}{2} \right) $

Given: y-intercept = -4, so:

$ \dfrac{7}{2} - \dfrac{(k - 1)(k + 1)}{2} = -4 $

Multiply both sides by 2:

$ 7 - (k^2 - 1) = -8 \Rightarrow 7 - k^2 + 1 = -8 \Rightarrow 8 - k^2 = -8 $

$ \Rightarrow k^2 = 16 \Rightarrow k = \pm4 $

✅ Final Answer: $\boxed{k = -4 \text{ or } 4}$

Qus : 12 NIMCET 2027 PYQ

Let $a$ be the distance between the lines $−2x + y = 2$ and $2x − y = 2$, and $b$ be the distance between the lines $4x − 3y= 5$ and $6y − 8x = 1$, then

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2022 PYQ

Solution

Qus : 13 NIMCET 2027 PYQ

Area of the parallelogram formed by the lines y=4x, y=4x+1, x+y=0 and x+y=1

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2022 PYQ

Solution

Qus : 14 NIMCET 2027 PYQ

The area enclosed within the curve |x|+|y|=2 is

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2022 PYQ

Solution

Qus : 15 NIMCET 2027 PYQ

A straight line through the point (4, 5) is such that its intercept between the axes is bisected at A, then its equation is

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2022 PYQ

Solution

Qus : 16 NIMCET 2027 PYQ

The points (1,1/2) and (3,-1/2) are

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2024 PYQ

Solution

Given:

Points: $ A = (1, \frac{1}{2}) $, $ B = (3, -\frac{1}{2}) $

Line: $ 2x + 3y = k $

Step 1: Evaluate $ 2x + 3y $

For A: $ 2(1) + 3\left(\frac{1}{2}\right) = \frac{7}{2} $
For B: $ 2(3) + 3\left(-\frac{1}{2}\right) = \frac{9}{2} $

✅ Option-wise Check:

In between the lines $ 2x + 3y = -6 $ and $ 2x + 3y = 6 $: ✔️ True since $ \frac{7}{2}, \frac{9}{2} \in (-6, 6) $
On the same side of $ 2x + 3y = 6 $: ✔️ True, both values are less than 6
On the same side of $ 2x + 3y = -6 $: ✔️ True, both values are greater than -6
On the opposite side of $ 2x + 3y = -6 $: ❌ False, both are on the same side

✅ Final Answer:

The correct statements are:

In between the lines $ 2x + 3y = -6 $ and $ 2x + 3y = 6 $
On the same side of the line $ 2x + 3y = 6 $
On the same side of the line $ 2x + 3y = -6 $

Qus : 17 NIMCET 2027 PYQ

The locus of the orthocentre of the triangle formed by the lines (1+p)x-py+p(1+p)=0, (1+p)(x-q)+q(1+ q)=0 and y=0 where p≠q is

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2018 PYQ

Solution

Straight Line

Qus : 18 NIMCET 2027 PYQ

Equation of the line perpendicular to x-2y=1 and passing through (1,1) is

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2018 PYQ

Solution

Qus : 19 NIMCET 2027 PYQ

Let $(x_0,y_0)\in \mathbb{Z}^2$ be a point on the straight line $8x-3y=11$ which is equidistant from the coordinate axes. Then, the point $(x_0,y_0)$ will lie only in:

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2026 PYQ

Solution

A point equidistant from the coordinate axes satisfies:

$|x|=|y|$

So, either

$y=x$

$y=-x$

Given line is:

$8x-3y=11$

Case 1:

$y=x$

Put $y=x$ in the line:

$8x-3x=11$

$5x=11$

$x=\frac{11}{5}$

This is not an integer, so this case is rejected.

Case 2:

$y=-x$

Put $y=-x$ in the line:

$8x-3(-x)=11$

$8x+3x=11$

$11x=11$

$x=1$

Then,

$y=-1$

So, the point is:

$(1,-1)$

Here $x>0$ and $y<0$, so the point lies in the IV quadrant.

Qus : 20 NIMCET 2027 PYQ

Let the line $\frac{x}{4}+\frac{y}{2}=1$ meets the x-axis and y-axis at A and B, respectively. M is the midpoint of side AB, and M' is the image of the point M across the line x + y = 1. Let the point P lie on the line x + y = 1 such that the $\Delta$ABP is an isosceles triangle with AP = BP. Then the distance between M' and P is

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2025 PYQ

Solution

A = $(4,0)$, B = $(0,2)$

Midpoint of AB: $M=\left(\frac{4+0}{2},\frac{0+2}{2}\right)=(2,1)$

Image of $M(2,1)$ across line $x+y=1$:

$M'=(0,-1)$

Since $AP=BP$, point $P$ lies on perpendicular bisector of AB.

Perpendicular bisector of AB passes through $M(2,1)$ and has slope $2$:

$y-1=2(x-2)$

$y=2x-3$

Also $P$ lies on: $x+y=1$

So, $x+2x-3=1$ $3x=4$

$x=\frac{4}{3}$ $y=1-\frac{4}{3}=-\frac{1}{3}$

So, $P=\left(\frac{4}{3},-\frac{1}{3}\right)$

Distance between $M'(0,-1)$ and $P$:

$d=\sqrt{\left(\frac{4}{3}-0\right)^2+\left(-\frac{1}{3}+1\right)^2}$

$d=\sqrt{\frac{16}{9}+\frac{4}{9}}$

$d=\sqrt{\frac{20}{9}}$

$d=\frac{2\sqrt5}{3}$

Final Answer: $\boxed{\frac{2\sqrt5}{3}}$

Qus : 21 NIMCET 2027 PYQ

If non-zero numbers a, b, c are in A.P., then the straight line ax + by + c = 0 always passes through a fixed point, then the point is

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2017 PYQ

Solution

Since a, b and c are in A. P. 2b = a + c

a –2b + c =0

The line passes through (1, –2).

Qus : 22 NIMCET 2027 PYQ

If the lines x + (a – 1)y + 1 = 0 and 2x + a²y – 1 = 0 are perpendicular, then the condition satisfies by a is

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2017 PYQ

Solution

Qus : 23 NIMCET 2027 PYQ

The lines $px+qy=1$ and $qx+py=1$ are respectively the sides AB, AC of the triangle ABC and the base BC is bisected at $(p,q)$. Equation of the median of the triangle through the vertex A is

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2021 PYQ

Solution

Qus : 24 NIMCET 2027 PYQ

Distance between the parallel lines $y = 2x + 4$ and $6x = 3y + 5$ is

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2010 PYQ

Solution

Rewrite second line: $6x - 3y - 5 = 0$ Divide by 3: $2x - y - \dfrac{5}{3} = 0$ First line: $y = 2x + 4 \Rightarrow 2x - y + 4 = 0$ Distance between parallel lines: $d = \dfrac{|c_2 - c_1|}{\sqrt{a^2 + b^2}}$ Here: $c_1 = 4$, $c_2 = -\dfrac{5}{3}$, $a=2,\ b=-1$ Compute: $d = \dfrac{\left|4 - \left(-\dfrac{5}{3}\right)\right|}{\sqrt{4+1}} = \dfrac{\left|\dfrac{12}{3}+\dfrac{5}{3}\right|}{\sqrt{5}} = \dfrac{17/3}{\sqrt{5}} = \dfrac{17\sqrt{5}}{15}$

Qus : 25 NIMCET 2027 PYQ

The obtuse angle between lines 2y = x + 1 and y = 3x + 2 is

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2025 PYQ

Solution

Given lines:

$2y = x + 1$ and $y = 3x + 2$.

From $2y = x + 1$

we get $y = \dfrac{1}{2}x + \dfrac{1}{2}$,

so slope $m_1 = \dfrac{1}{2}$.

From $y = 3x + 2$,

slope $m_2 = 3$.

Angle between two lines: $ \tan\theta = \left|\dfrac{m_2 - m_1}{1 + m_1 m_2}\right| $

$ \tan\theta = \left|\dfrac{3 - \dfrac{1}{2}}{1 + \dfrac{1}{2}\cdot 3}\right| = \dfrac{\dfrac{5}{2}}{\dfrac{5}{2}} = 1 $

So $\theta = 45^\circ$ or $135^\circ$.

Required obtuse angle $= 135^\circ$.

Qus : 26 NIMCET 2027 PYQ

The foot of the perpendicular from the point (2, 4) upon $x + y = 1$ is

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2015 PYQ

Solution

Qus : 27 NIMCET 2027 PYQ

If $(– 4, 5)$ is one vertex and $7x – y + 8 = 0$ is one diagonal of a square, then the equation of the other diagonal is

1
2
3
4
Go to Discussion

NIMCET 2027 Previous Year PYQ NIMCET 2027 NIMCET 2015 PYQ

Solution

NIMCET 2027

Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

NIMCET 2027

Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

Ask Your Question or Put Your Review.

Please provide more information - at least 10 characters

Aspire's Library

NIMCET 2027 Previous Year Questions (PYQs)

NIMCET 2027 Straight Line PYQ

Solution

Solution

Same Side of a Line — Geometric Condition

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Given: Points: \( P(1, 4) \), \( Q(k, 3) \)

Solution

Solution

Solution

Solution

Solution

Given:

Step 1: Evaluate \( 2x + 3y \)

✅ Option-wise Check:

✅ Final Answer:

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

NIMCET 2027

NIMCET 2027

Feedback

Info.

Information

Aspire's Library

NIMCET 2027 Previous Year Questions (PYQs)

NIMCET 2027 Straight Line PYQ

Solution

Solution

Same Side of a Line — Geometric Condition

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Given: Points: \( P(1, 4) \), \( Q(k, 3) \)

Solution

Solution

Solution

Solution

Solution

Given:

Step 1: Evaluate \( 2x + 3y \)

✅ Option-wise Check:

✅ Final Answer:

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

NIMCET 2027

NIMCET 2027

Game Changer NIMCET Test Series 2026

Info.

Confirm Delete

Edit