NIMCET Straight Line Previous Year Questions (PYQs) – Page 3 of 3
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Showing 27 questions
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
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
1
2
3
4
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
1
2
3
4
Solution
Distance between the parallel lines
$y = 2x + 4$
and
$6x = 3y + 5$
is
1
2
3
4
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:
$d = \dfrac{\left|4 - \left(-\dfrac{5}{3}\right)\right|}{\sqrt{4+1}}
= \dfrac{\left|\dfrac{12}{3}+\dfrac{5}{3}\right|}{\sqrt{5}}
= \dfrac{17/3}{\sqrt{5}}
= \dfrac{17\sqrt{5}}{15}$
The obtuse angle between lines 2y = x + 1 and y = 3x + 2 is
1
2
3
4
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 $= 135^\circ$.
The foot of the perpendicular from the point (2, 4) upon $x + y = 1$ is
1
2
3
4
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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2
3
4
Solution
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