Phrases Straight line Previous Year Question Paper and Solution with PDF

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Phrases Previous Year Questions (PYQs)

Phrases 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 $2x + 3y = 6$ : 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

$(– 4, 5)$ 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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Phrases Previous Year Questions (PYQs)

Phrases Straight Line PYQ

Solution

Solution

Same Side of a Line — Geometric Condition

Solution

Solution

Solution

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

Solution

Solution

Solution

Solution

Solution

Given:

Step 1: Evaluate 2x+3y 2x + 3y

✅ Option-wise Check:

✅ Final Answer:

Solution

Solution

Solution

Solution

Solution

Solution

Solution

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

Step 1: Evaluate $2x + 3y$