NIMCET Straight line Previous Year Question Paper and Solution with PDF

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Previous Year Questions
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NIMCET Previous Year Questions (PYQs)

NIMCET Straight Line PYQ

Qus : 1

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

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Solution

Qus : 2

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

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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

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

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Solution

Qus : 4

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

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Solution

Qus : 5

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

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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 : 6

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

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Solution

Qus : 7

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

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Solution

Qus : 8

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

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Solution

Qus : 9

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

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Solution

Qus : 10

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

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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 : 11

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

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Solution

Straight Line

Qus : 12

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

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Solution

Qus : 13

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

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Solution

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

a –2b + c =0

The line passes through (1, –2).

Qus : 14

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

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Solution

Qus : 15

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

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Solution

Qus : 16

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

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Solution

Qus : 17

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

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Solution

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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

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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

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