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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





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Solution



The range of values of θ in the interval (0,π) such that the points (3, 2) and (cosθ,sinθ) 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+y1=0
Point 1: (3, 2) → Lies on the side where value is positive:
f(3,2)=3+21=4>0

Point 2: (cosθ,sinθ) lies on same side if: cosθ+sinθ>1 Using identity: cosθ+sinθ=2sin(θ+π4)sin(θ+π4)>12 So: θ(0, π2)

✅ Final Answer: θ(0, π2)



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



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



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 = (1+k2,4+32)=(1+k2,72)

Step 2: Find slope of PQ

Slope of PQ = 34k1=1k1

Step 3: Slope of perpendicular bisector = negative reciprocal = k1

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

y72=(k1)(x1+k2)

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

y=72+(k1)(1+k2)

y=72(k1)(1+k2)

Given: y-intercept = -4, so:

72(k1)(k+1)2=4

Multiply both sides by 2:

7(k21)=87k2+1=88k2=8

k2=16k=±4

✅ Final Answer: k=4 or 4



Let a be the distance between the lines 2x+y=2 and 2xy=2, and b be the distance between the lines 4x3y=5 and 6y8x=1, then





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Solution



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



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





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Solution



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



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





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Solution

Given:

Points: A=(1,12), B=(3,12)

Line: 2x+3y=k

Step 1: Evaluate 2x+3y

For A: 2(1)+3(12)=72
For B: 2(3)+3(12)=92

✅ Option-wise Check:

  • In between the lines 2x+3y=6 and 2x+3y=6: ✔️ True since 72,92(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:

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


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


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





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Solution



Let the line x4+y2=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 ΔABP is an isosceles triangle with AP = BP. Then the distance between M' and P is





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Solution



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).


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



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



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





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Solution



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





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Solution



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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