nimcet 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
Previous Year Questions

nimcet
Previous Year Questions
nimcet

nimcet Previous Year Questions (PYQs)

nimcet Straight Line 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 Previous Year PYQnimcet NIMCET 2023 PYQ

Solution

Qus : 2nimcet PYQ

The range of values of $\theta$ in the interval $(0,\pi)$ such that the points (3, 2) and $(cos\theta ,sin\theta)$ lie on the samesides of the line x + y – 1 = 0, is

1
2
3
4
Go to Discussion

nimcet Previous Year PYQnimcet 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 : 3nimcet 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 Previous Year PYQnimcet NIMCET 2019 PYQ

Solution

Qus : 4nimcet 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 Previous Year PYQnimcet NIMCET 2019 PYQ

Solution

Qus : 5nimcet 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 Previous Year PYQnimcet 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 : 6nimcet 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 Previous Year PYQnimcet NIMCET 2022 PYQ

Solution

Qus : 7nimcet 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 Previous Year PYQnimcet NIMCET 2022 PYQ

Solution

Qus : 8nimcet PYQ

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

1
2
3
4
Go to Discussion

nimcet Previous Year PYQnimcet NIMCET 2022 PYQ

Solution

Qus : 9nimcet 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 Previous Year PYQnimcet NIMCET 2022 PYQ

Solution

Qus : 10nimcet PYQ

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

1
2
3
4
Go to Discussion

nimcet Previous Year PYQnimcet 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 : 11nimcet 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 Previous Year PYQnimcet NIMCET 2018 PYQ

Solution

Straight Line

Qus : 12nimcet PYQ

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

1
2
3
4
Go to Discussion

nimcet Previous Year PYQnimcet NIMCET 2018 PYQ

Solution

Qus : 13nimcet 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 Previous Year PYQnimcet 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 : 14nimcet 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 Previous Year PYQnimcet NIMCET 2017 PYQ

Solution

Qus : 15nimcet 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 Previous Year PYQnimcet NIMCET 2021 PYQ

Solution

Qus : 16nimcet PYQ

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

1
2
3
4
Go to Discussion

nimcet Previous Year PYQnimcet NIMCET 2015 PYQ

Solution

Qus : 17nimcet 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 Previous Year PYQnimcet NIMCET 2015 PYQ

Solution

nimcet

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

Click Here to
View More

nimcet

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 Previous Year Questions (PYQs)

nimcet Straight Line PYQ

Solution

Solution

Same Side of a Line — Geometric Condition

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

nimcet

nimcet

Feedback

Info.

Information

Aspire's Library

nimcet Previous Year Questions (PYQs)

nimcet Straight Line PYQ

Solution

Solution

Same Side of a Line — Geometric Condition

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

nimcet

nimcet

Info.

Confirm Delete

Edit