Two Dimensional Geometry jee mains questions | JEE Main Two Dimensional Geometry Previous Year Question

Q 1 :

The vertices of a triangle are A(–1, 3), B(–2, 2) and C(3, –1). A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is [2024]

$- x + y - (2 - \sqrt{2}) = 0$
$x - y - (2 + \sqrt{2}) = 0$
$x + y + (2 - \sqrt{2}) = 0$
$x + y - (2 - \sqrt{2}) = 0$

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(4)

Equation of AC : $y - 3 = \frac{- 4}{4} (x + 1)$

$\Rightarrow x + y = 2$

Equation AB : $y - 2 = \frac{1}{1} (x + 2)$

$\Rightarrow x - y + 4 = 0$

Equation BC : $y + 1 = \frac{- 3}{5} (x - 3)$

$\Rightarrow 3 x + 5 y = 4$

Distance of AC from origin O(0, 0) = $\frac{| - 2 |}{\sqrt{2}} = \sqrt{2}$

Distance of AB from origin O(0, 0) = $\frac{| 4 |}{\sqrt{2}} = 2 \sqrt{2}$

Distance of BC from origin O(0, 0) = $\frac{| - 4 |}{\sqrt{9 + 25}} = \frac{4}{\sqrt{34}}$

When each side of $∆ A B C$ shifted by one unit inwards distance of new sides from origin are $\sqrt{2} - 1, 2 \sqrt{2} - 1, \frac{4}{\sqrt{34}} + 1$ respectively.

Clearly new position of AC is nearest to the origin.

Equation of AC when shifted by one unit inwareds can be chosen as x + y = c

Thus, we have $\frac{| c - 2 |}{\sqrt{2}} = 1 \Rightarrow c = 2 - \sqrt{2}$

$∴$ Required equation is $x + y - 2 + \sqrt{2} = 0$ .

Q 2 :

Let two straight lines drawn from the origin O intersect the line 3x + 4y = 12 at the points P and Q such that $△ O P Q$ is an isosceles triangle and $∠ P O Q = 90 °$ . If $l = O P^{2} + P Q^{2} + Q O^{2}$ , then the greatest integer less than or equal to $l$ is: [2024]

44
42
46
48

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(3)

Let $P (r \cos θ, r \sin θ)$ and $Q (- r \sin θ, r \cos θ)$

$∴ O P^{2} = O Q^{2} = r^{2}$

Now, in $△ P O Q, ∠ P O Q = 90 °, P Q^{2} = 2 r^{2}$

Now, $l = O P^{2} + P Q^{2} + O Q^{2} = 4 r^{2}$

$∵$ P and Q lies on 3x + 4y = 12

$∴ 3 (r \cos θ) + 4 (r \sin θ) = 12$

$\Rightarrow 3 \cos θ + 4 \sin θ = \frac{12}{r}$ ... (i)

and $3 (- r \sin θ) + 4 (r \cos θ) = 12$

$\Rightarrow - 3 \sin θ + 4 \cos θ = \frac{12}{r}$ ... (ii)

From (i) and (ii), we have

$25 = \frac{288}{r^{2}} \Rightarrow r^{2} = \frac{288}{25}$

Now, $l = 4 r^{2} = 4 \times \frac{288}{25} = 46.08$

So, $[l] = 46$ .

Q 3 :

Let A(–1, 1) and B(2, 3) be two points and P be a variable point above the line AB such that the area of $△ P A B$ is 10. If the locus of P is ax + by = 15, then 5a + 2b is: [2024]

6
4
$- \frac{6}{5}$
$- \frac{12}{5}$

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(4)

Let coordinates of P be (h, k).

Area of $△ P A B$ = 10sq. units

$\Rightarrow 10 = \frac{1}{2}$ $| \begin{matrix} h \\ - 1 \\ 2 \end{matrix} \begin{matrix} k \\ 1 \\ 3 \end{matrix} \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} |$

$\Rightarrow \pm 20 = h (- 2) - k (- 3) + 1 (- 5)$

$\Rightarrow - 2 h + 3 k = 25$

( $∵$ P is above the line AB, so we take only +ve value)

$\Rightarrow \frac{- 2 \times 3}{5} h + \frac{3 \times 3}{5} k = \frac{3 \times 25}{5}$

$\Rightarrow \frac{- 6}{5} h + \frac{9}{5} k = 15$

On comparing it with ax + by = 15, we get $a = \frac{- 6}{5}$ and $b = \frac{9}{5}$

. $5 a + 2 b = - 6 + \frac{18}{5} = \frac{- 12}{5}$

Q 4 :

The equations of two sides AB and AC of a triangle ABC are 4x + y = 14 and 3x – 2y = 5, respectively. The point $(2, - \frac{4}{3})$ divides the third side BC internally in the ratio 2 : 1, the equation of the side BC is [2024]

x + 3y + 2 = 0
x – 3y – 6 = 0
x + 6y + 6 = 0
x – 6y – 10 = 0

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(1)

Let $D (2, \frac{- 4}{3})$ be the point on BC which divides BC in ratio 2 : 1.

Let coordinates of B be $(x_{1}, 14 - 4 x_{1})$ and coordinates of C be $(x_{2}, \frac{3 x_{2} - 5}{2})$

Now, $\frac{2 x_{2} + x_{1}}{3} = 2$ , $\frac{2 (\frac{3 x_{2} - 5}{2}) + 14 - 4 x_{1}}{3} = \frac{- 4}{3}$

$\Rightarrow 2 x_{2} + x_{1} = 6, 3 x_{2} - 4 x_{1} = - 4 + 5 - 14$

$\Rightarrow 2 x_{2} + x_{1} = 6, 3 x_{2} - 4 x_{1} = - 13$

On solving these two equations, we get, $x_{1} = 4, x_{2} = 1$

So B(4, –2) and C(1, –1) are the required points.

Equation of BC : $y + 1 = \frac{- 1}{3} (x - 1) \Rightarrow 3 y + x + 2 = 0$

Q 5 :

The Portion of the line 4x + 5y = 20 in the first quadrant is trisected by the lines $L_{1}$ and $L_{2}$ passing through the origin. The tangent of an angle between the lines $L_{1}$ and $L_{2}$ is: [2024]

$\frac{30}{41}$
$\frac{25}{41}$
$\frac{2}{5}$
$\frac{8}{5}$

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(1)

Lines $L_{1}$ and $L_{2}$ trisect the line 4x + 5y = 20.

$x_{1} = \frac{5 \times 1 + 0 \times 2}{1 + 2} = \frac{5}{3}$

$y_{1} = \frac{0 \times 1 + 4 \times 2}{1 + 2} = \frac{8}{3}$

$(x_{1}, y_{1}) \equiv (\frac{5}{3}, \frac{8}{3})$

Similarly,

$x_{2} = \frac{0 \times 1 + 5 \times 2}{1 + 2} = \frac{10}{3}$

$y_{2} = \frac{4 \times 1 + 0 \times 2}{1 + 2} = \frac{4}{3}, (x_{2}, y_{2}) \equiv (\frac{10}{3}, \frac{4}{3})$

Slope of line $L_{1} : m_{1} = \frac{8}{3} \times \frac{3}{5} = \frac{8}{5}$

Slope of line $L_{2} : m_{2} = \frac{4}{3} \times \frac{3}{10} = \frac{2}{5}$

Tangent angle between the lines $L_{1}$ and $L_{2}$ :

$\tan θ =$ $| \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} |$ $=$ $| \frac{\frac{8}{5} - \frac{2}{5}}{1 + \frac{8}{5} \times \frac{2}{5}} |$ $= \frac{30}{41}$

Q 6 :

Let R be the interior region between the lines 3x – y + 1 = 0 and x + 2y – 5 = 0 containing the origin. The set of all values of a, for which the points $(a^{2}, a + 1)$ lie in R, is: [2024]

$(- 3, - 1) \cup (\frac{1}{3}, 1)$
$(- 3, 0) \cup (\frac{2}{3}, 1)$
$(- 3, - 1) \cup (- \frac{1}{3}, 1)$
$(- 3, 0) \cup (\frac{1}{3}, 1)$

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(4)

It is given that, region R lies between the lines 3x – y + 1 = 0 and x + 2y – 5 = 0.

The point $(a^{2}, a + 1)$ and (0, 0) lie in the region R.

$\Rightarrow (a^{2}, a + 1)$ and (0, 0) are on same side of both the line.

For Line 3x – y + 1 = 0, O(0, 0) is on the right side of the line.

So, point $(a^{2}, a + 1)$ will also be on right side of the line.

$\Rightarrow 3 a^{2} - a - 1 + 1 > 0$

$\Rightarrow a (3 a - 1) > 0$

$\Rightarrow a \in (- \infty, 0) \cup (\frac{1}{3}, \infty)$ ... (i)

For line x + 2y – 5 = 0, O(0, 0) is on the left side of the line.

So, point $(a^{2}, a + 1)$ will also be on left side of the line.

$\Rightarrow a^{2} + 2 a + 2 - 5 < 0$

$\Rightarrow (a + 3) (a - 1) < 0$

$\Rightarrow a \in (- 3, 1)$ ... (ii)

From the intersection of (i) and (ii), we get

$a \in (- 3, 0) \cup (\frac{1}{3}, 1)$

Q 7 :

In a $△ A B C$ , Suppose y = x is the equation of the bisector of the angle B and the equation of the side AC is 2x – y = 2. If 2AB = BC and the points A and B are respectively (4, 6) and $(α, β)$ , then $α + 2 β$ is equal to [2024]

39
42
48
45

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(2)

We have, y = x ... (i)

and 2x – y = 2 ... (ii)

Solving (i) and (ii), we get x = 2 and y = 2

As, $△ A B D \tilde{=} △ C B D$

$∴ \frac{A B}{B C} = \frac{A D}{D C} = \frac{1}{2}$

$D (\frac{λ + 8}{3}, \frac{2 λ - 2 + 12}{3}) lies on y = x$

$∴ \frac{λ + 8}{3} = \frac{2 λ + 10}{3} \Rightarrow λ = - 2 . S o, C \equiv (- 2, - 6)$

The image of the point A will lies on the line BC.

Let A' = (6, 4)

Now, $\frac{β - 4}{α - 6} = \frac{10}{8} \Rightarrow \frac{α - 4}{α - 6} = \frac{5}{4} (∵ α = β)$

$\Rightarrow 4 α - 16 = 5 α - 30 \Rightarrow 14 = α = β$

$∴ α + 2 β = 14 + 2 \times 14 = 42$

Q 8 :

The distance of the point (2, 3) from the line 2x – 3y + 28 = 0, measured parallel to the line $\sqrt{3} x - y + 1 = 0$ , is equal to [2024]

$4 + 6 \sqrt{3}$
$4 \sqrt{2}$
$6 \sqrt{3}$
$3 + 4 \sqrt{2}$

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(1)

Slope of a line passing through (2, 3) and parallel to line $\sqrt{3} x - y + 1 = 0$ is same as that of line $\sqrt{3} x - y + 1 = 0$

So, slope of line = $\sqrt{3}$

$\Rightarrow \tan θ = \sqrt{3} \Rightarrow \sin θ = \frac{\sqrt{3}}{2}, \cos θ = \frac{1}{2}$

So, equation of line passing through (2, 3) and having slope $\sqrt{3}$ in normal form is,

$\frac{x - 2}{1 / 2} = \frac{y - 3}{\sqrt{3} / 2} = r \Rightarrow x = \frac{r}{2} + 2, y = \frac{\sqrt{3}}{2} r + 3$

This line passes through 2x – 3y + 28 = 0

$∴ 2 (\frac{r}{2} + 2) - 3 (\frac{\sqrt{3}}{2} r + 3) + 28 = 0 \Rightarrow (\frac{2 - 3 \sqrt{3}}{2}) r = - 23$

$\Rightarrow r = \frac{- 23 \times 2}{2 - 3 \sqrt{3}} \times \frac{2 + 3 \sqrt{3}}{2 + 3 \sqrt{3}} \Rightarrow r = 2 (2 + 3 \sqrt{3}) = 4 + 6 \sqrt{3}$

Q 9 :

Let A be the point of intersection of the lines 3x + 2y = 14, 5x – y = 6 and B be the point of intersection of the lines 4x + 3y = 8, 6x + y = 5. The distance of the point P(5, –2) from the line AB is [2024]

8
$\frac{5}{2}$
$\frac{13}{2}$
6

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(4)

Point of intersection of lines 3x + 2y = 14 and 5x – y = 6 is A(2, 4)

Point of intersection of lines 4x + 3y = 8 and 6x + y = 5 is $B = (\frac{1}{2}, 2)$

So, equation of line AB is, $y - 4 = \frac{2 - 4}{\frac{1}{2} - 2} (x - 2)$

$\Rightarrow y - 4 = \frac{4}{3} (x - 2) \Rightarrow 4 x - 3 y + 4 = 0$

$∴$ Distance of point P(5, –2) from 4x – 3y + 4 is given as

$d =$ $| \frac{4 \times 5 - 3 \times (- 2) + 4}{\sqrt{16 + 9}} |$ $\Rightarrow d = 6 units$

Q 10 :

A line passing through the point A(9, 0) makes an angle of 30° with the positive direction of x-axis. If this line is rotated about A through an angle of 15° in the clockwise direction, then its equations in the new position is [2024]

$\frac{x}{\sqrt{3} + 2} + y = 9$
$\frac{y}{\sqrt{3} - 2} + x = 9$
$\frac{x}{\sqrt{3} - 2} + y = 9$
$\frac{y}{\sqrt{3} + 2} + x = 9$

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(2)

Let $l$ be the line passing through point A(9, 0) making angle 30° with x-axis.

Line is rotated clockwise by 15° then $l'$ is the new position of line where it make 15° angle with x-axis.

So, equation of line passing through (9, 0) and making angle 15° with x-axis is (y – 0) = tan 15° (x – 9)

$\Rightarrow$ y = tan (45° – 30°)(x – 9) = $(\frac{\tan 45 ° - \tan 30 °}{1 + \tan 45 ° \cdot \tan 30 °}) (x - 9)$

$\Rightarrow y = (2 - \sqrt{3}) (x - 9)$

$\Rightarrow \frac{y}{2 - \sqrt{3}} = x - 9 \Rightarrow \frac{y}{\sqrt{3} - 2} + x = 9$

Q 11 :

If $x^{2} - y^{2} + 2 h x y + 2 g x + 2 f y + c = 0$ is the locus of a point, which moves such that it is always equidistant from the lines x + 2y + 7 = 0 and 2x – y + 8 = 0, then the value of g + c + h – f equals [2024]

29
6
8
14

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(4)

Let point P(x, y) be equidistant from the given lines.

$∴ \frac{x + 2 y + 7}{\sqrt{5}} = \pm \frac{2 x - y + 8}{\sqrt{5}}$

$\Rightarrow {(x + 2 y + 7)}^{2} = {(2 x - y + 8)}^{2}$

$\Rightarrow x^{2} + 4 y^{2} + 49 + 4 x y + 28 y + 14 x = 4 x^{2} + y^{2} + 64 - 4 x y - 16 y + 32 x$

$\Rightarrow 3 x^{2} - 3 y^{2} - 8 x y + 18 x - 44 y + 15 = 0$

$\Rightarrow x^{2} - y^{2} - \frac{8}{3} x y + 6 x - \frac{44}{3} y + 5 = 0$ ... (i)

This is the locus of the point P(x, y).

Now, compare equation (i) with given equation of locus, we get

$2 h = \frac{- 8}{3} \Rightarrow h = \frac{- 4}{3}$ ,

$2 g = 6 \Rightarrow g = 3$ ,

$2 f = \frac{- 44}{3} \Rightarrow \frac{- 22}{3}$ ,

and c = 5

$∴ g + c + h - f = 3 + 5 - \frac{4}{3} + \frac{22}{3} = 14$ .

Q 12 :

Let A (a, b), B(3, 4) and C(–6, –8) respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point P(2a + 3, 7b + 5) from the line 2x + 3y – 4 = 0 measured parallel to the line x – 2y – 1 = 0 is [2024]

$\frac{17 \sqrt{5}}{6}$
$\frac{\sqrt{5}}{17}$
$\frac{17 \sqrt{5}}{7}$
$\frac{15 \sqrt{5}}{7}$

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(3)

Given the A (a, b), B(3, 4) and C(–6, –8) respectively the centroid, circumcentre and orthocentre of a triangle.

We know that centroid divides circumcentre and orthocentre internally in the ratio 1 : 2

$∴ a = \frac{- 6 + 6}{2}, b = \frac{- 8 + 8}{3}$

$\Rightarrow$ a = 0, b = 0

Therefore, the coordinates of P are (3, 5).

Also, $l$ : 2x + 3y – 4 = 0 (Given) ... (i)

Equation of the line passing through P(3, 5) and parallel to the line x – 2y – 1 = 0 is

$y - 5 = \frac{1}{2} (x - 3)$

$\Rightarrow 2 y - 10 = x - 3$

$\Rightarrow x - 2 y + 7 = 0$ ... (ii)

Solving equation (i) and (ii), we get

$x = \frac{- 13}{7}, y = \frac{18}{7}$

$∴$ Distance between (3, 5) and $(\frac{- 13}{7}, \frac{18}{7})$ is

$\sqrt{{(3 + \frac{13}{7})}^{2} + {(5 - \frac{18}{7})}^{2}} = \sqrt{\frac{1156}{49} + \frac{289}{49}} \sqrt{\frac{1445}{49}} = \frac{17 \sqrt{5}}{7}$

Q 13 :

Consider a triangle ABC having the vertices A(1, 2), $B (α, β)$ and $C (γ, δ)$ and angles $∠ A B C = \frac{π}{6}$ and $∠ B A C = \frac{2 π}{3}$ . If the points B and C lie on the line y = x + 4, then $α^{2} + γ^{2}$ is equal to __________ [2024]

(14)

$P = \frac{| - 1 + 2 - 4 |}{\sqrt{1^{2} + 1^{2}}} = \frac{3}{\sqrt{2}}$

$\sin (\frac{π}{6}) = \frac{3 / \sqrt{2}}{A B} \Rightarrow \frac{3}{\sqrt{2} A B} = \frac{1}{2} \Rightarrow A B = \frac{6}{\sqrt{2}}$

$\Rightarrow {(α - 1)}^{2} + {(α + 4 - 2)}^{2} = {(\frac{6}{\sqrt{2}})}^{2}$

$\Rightarrow α^{2} + 1 - 2 α + α^{2} + 4 + 4 α = 18$

$\Rightarrow 2 α^{2} + 2 α - 13 = 0 \to α and γ$ satisfy same equation

$∴$ $α$ and $γ$ are the roots of the equation $2 x^{2} + 2 x - 13 = 0$

Sum of roots = $α + γ = \frac{- 2}{2} = - 1$

Product of roots = $α γ = \frac{- 13}{2}$

Now, $α^{2} + γ^{2} = {(α + γ)}^{2} - 2 α γ$

$= {(- 1)}^{2} - 2 (\frac{- 13}{2}) = 1 + 13 = 14$ .

Q 14 :

Let a ray of light passing through the point (3, 10) reflects on the line 2x + y = 6 and the reflected ray passes through the point (7, 2). If the equation of the incident ray is ax + by + 1 = 0, then $a^{2} + b^{2} + 3 a b$ is equal to __________ . [2024]

(1)

$\frac{x - 3}{2} = \frac{y - 10}{1} = \frac{- 2 (2 \times 3 + 10 - 6)}{4 + 1} = - 4$

$\Rightarrow x = - 5, y = 6$

$∴$ A'(–5, 6) and B(7, 2)

$∴$ Equation of line A'B is

$\Rightarrow y - 6 = \frac{2 - 6}{7 + 5} (x + 5)$

$\Rightarrow y - 6 = - \frac{1}{3} (x + 5)$

$\Rightarrow 3 y - 18 = - x - 5 \Rightarrow x + 3 y = 13$

and 2x + y = 6 (Given line).

On solving, we get y = 4, x = 1

$∴ Q \equiv (1, 4)$

Equation of line AQ is

$y - 10 = \frac{10 - 4}{3 - 1} (x - 3) \Rightarrow y - 10 = 3 (x - 3)$

$\Rightarrow y - 10 = 3 x - 9 \Rightarrow 3 x - y + 1 = 0$

On comparing with given equation ax + by + 1 = 0, we get a = 3, b = –1

Hence, $a^{2} + b^{2} + 3 a b = 9 + 1 + 3 (3) (- 1) = 9 + 1 - 9 = 1$ .

Q 15 :

Let ABC be an isosceles triangle in which A is at (–1, 0), $∠ A = \frac{2 π}{3}$ , AB = AC and B is on the positive x-axis. If $B C = 4 \sqrt{3}$ and the line BC intersects the line y = x + 3 at $(α, β)$ , then $\frac{β^{4}}{α^{2}}$ is __________. [2024]

(36)

We have, $\frac{\sin 30 °}{A C} = \frac{\sin 120 °}{B C}$

$\Rightarrow A C = 4 \sqrt{3} \times \frac{1}{2} \times \frac{2}{\sqrt{3}}$

$\Rightarrow A C = 4 = A B$

$\Rightarrow A B^{2} = {(x + 1)}^{2} \Rightarrow x + 1 = \pm 4$

But B is on positive x-axis

$∴$ Coordinates of B = (3, 0)

Now, $A C^{2} = 16$

$\Rightarrow {(p + 1)}^{2} + q^{2} = 16$ ... (i)

Also, $B C^{2} = 48$

$\Rightarrow {(p - 3)}^{2} + q^{2} = 48$ ... (ii)

Solving (i) and (ii), we get

$p = - 3, q = 2 \sqrt{3}$

So, coordinates of $C = (- 3, 2 \sqrt{3})$

$∴$ Equation of line BC is, $y - 0 = \frac{2 \sqrt{3} - 0}{- 3 - 3} (x - 3)$

$\Rightarrow x + \sqrt{3} y = 3$ ... (iii)

Now, point of intersection of line (iii) and x + 3 = y is

$(3 (\sqrt{3} - 2), 3 (\sqrt{3} - 1))$

$\Rightarrow α = 3 (\sqrt{3} - 2), β = 3 (\sqrt{3} - 1)$

$∴$ $\frac{β^{4}}{α^{2}} = \frac{3^{4} {(3 + 1 - 2 \sqrt{3})}^{2}}{3^{2} {(\sqrt{3} - 2)}^{2}} = \frac{9 (2 {(2 - \sqrt{3}))}^{2}}{{(2 - \sqrt{3})}^{2}} = 9 \times 4 = 36$ .

Q 16 :

If the sum of squares of all real values of $α$ , for which the lines 2x – y + 3 = 0, 6x + 3y + 1 = 0 and $α x + 2 y - 2 = 0$ do not form a triangle is p, then the greatest integer less than or equal to p is __________. [2024]

(32)

We have, 2x – y + 3 = 0 ... (i)

6x + 3y + 1 = 0 ... (ii)

$α x + 2 y - 2 = 0$ ... (iii)

Case I. If the lines are concurrent then they do not form a triangle

$∴$ $| \begin{matrix} 2 & - 1 & 3 \\ 6 & 3 & 1 \\ α & 2 & - 2 \end{matrix} |$ $= 0$

$\Rightarrow α (- 1 - 9) - 2 (2 - 18) - 2 (6 + 6) = 0$

$\Rightarrow - 10 α + 32 - 24 = 0 \Rightarrow 10 α = 8 \Rightarrow α = \frac{4}{5}$

Case II. If the lines are parallel then they do not form a triangle.

If the lines (i) and (iii) are parallel,

$∴ \frac{2}{α} = \frac{- 1}{2} \neq \frac{- 3}{2} \Rightarrow α = - 4$

If the lines (ii) and (iii) are parallel,

$∴ \frac{6}{α} = \frac{3}{2} \neq \frac{- 1}{2} \Rightarrow α = 4$

$∴$ Sum of squares of all real values of $α$

$= {(\frac{4}{5})}^{2} + {(- 4)}^{2} + {(4)}^{2} = \frac{16}{25} + 16 + 16 = \frac{16}{25} + 32$

$[P] = [32 + \frac{16}{25}] = 32$ .

Q 17 :

Let for any three distinct consecutive terms a, b, c of an A.P., the lines ax + by + c = 0 be concurrent at the point P and $Q (α, β)$ be a point such that the system of equations x + y + z = 6, 2x + 5y + $α$ z = $β$ and x + 2y + 3z = 4, has infinitely many solutions. Then ${(P Q)}^{2}$ is equal to __________. [2024]

(113)

a, b, c are in A.P. $\Rightarrow$ 2b = a + c $\Rightarrow$ a – 2b + c = 0 and ax + by + c = 0 are concurrent.

Two lines $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ are concurrent if

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} ∴ \frac{1}{x} = \frac{- 2}{y} = \frac{1}{1} = 0$

$\Rightarrow x = 1 and y = - 2$

So, P(1, –2) is the point of concurrency.

Now, x + y + z = 6 ... (i)

2x + 5y + $α$ z = $β$ ... (ii)

x + 2y + 3z = 4 ... (iii)

has infinitely many solution.

On solving these equation, we have

x + y = 6 – z, x + 2y = 4 – 3z

On solving these two, we get –y = 2 + 2z

$\Rightarrow$ y = –2(1 + z) $\Rightarrow$ x = 6 – z + 2 + 2z $\Rightarrow$ x = 8 + z

From (ii), we get 2(8 + z) + 5(–2(1 + z)) + $α$ z = $β$

$\Rightarrow 6 - 8 z + α z = β$

$\Rightarrow z (α - 8) = β - 6$

For infinitely many solution, $α = 8, β = 6$

$| P Q |^{2}$ $= {(\sqrt{{(8 - 1)}^{2} + {(6 + 2)}^{2}})}^{2} = 113$

Q 18 :

Let the area of the triangle formed by a straight line L : x + by + c = 0 with co-ordinate axes be 48 square units. If the perpendicular drawn from the origin to the line L makes an angle of 45° with the positive x-axis, then the value of $b^{2} + c^{2}$ is : [2025]

83
97
93
90

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(2)

We have, area of $△$ OAB = 48 sq. units

Now, slope of line OD is m = tan 45° = 1

$∴$ Slope of line $A B = \frac{- 1}{m} = - 1$ ... (i)

Also, equation of line AB is x + by + c = 0

So, Slope $\frac{- 1}{b}$ ... (ii)

From (i) and (ii), we get

$\Rightarrow - 1 = \frac{- 1}{b} \Rightarrow b = 1$

Now, x and y intercept of line x + by + c = 0 are given as x = – c and $y = \frac{- c}{b}$ respectively.

$∴$ Area of $△$ OAB = $\frac{1}{2} \times (- c) \times \frac{- c}{b} \Rightarrow 48 = \frac{c^{2}}{2 b}$

$\Rightarrow c^{2} = 48 \times 2$ [ $∵$ b = 1]

$\Rightarrow c^{2} = 96$

So, $b^{2} + c^{2} = 1 + 96 = 97$ .

Q 19 :

A line passes through the origin and makes equal angles with the positive coordinate axes. It intersects the lines $L_{1} : 2 x + y + 6 = 0$ and $L_{2} : 4 x + 2 y - p = 0$ , p > 0, at the points A and B, respectively. If $A B = \frac{9}{\sqrt{2}}$ and the foot of the perpendicular from the point A on the line $L_{2}$ is M, then $\frac{A M}{B M}$ is equal to [2025]

4
3
5
2

1

2

3

4

(2)

We have, $L_{1} : 2 x + y + 6 = 0$ and $L_{2} : 4 x + 2 y - p = 0$

Line y = x passes through origin and makes equal angles with the positive coordinates axes.

Slope of y = x is $m_{1} = 1$ , Slope of $L_{2}$ is $m_{2} = - 2$

So, $\tan θ =$ $| \frac{1 + 2}{1 - 2} |$ $= 3$

Now, in $△$ ABM

$\Rightarrow \tan θ = \frac{A M}{B M} = 3$ .

Q 20 :

The shortest distance between the curves $y^{2} = 8 x$ and $x^{2} + y^{2} + 12 y + 35 = 0$ is: [2025]

$3 \sqrt{2} - 1$
$2 \sqrt{3} - 1$
$\sqrt{2}$
$2 \sqrt{2} - 1$

1

2

3

4

(4)

Given, $x^{2} + y^{2} + 12 y + 35 = 0$ , which is a circle having centre, C = (0, –6)

Radius, $r = \sqrt{36 - 35} = 1$

Since, the shortest distance between both curves will be normal to the curve $y^{2} = 8 x$ from centre (0, –6), as shown in figure.

$∴$ Equation of normal to the parabola

$y^{2} = 8 x$ is $y = m x - 4 m - 2 m^{3}$

where m is slope, passes through (0, –6), then –6 = –4m – $2 m^{3}$

$\Rightarrow m^{3} + 2 m - 3 = 0$

$\Rightarrow (m - 1) (m^{2} + m + 3) = 0$

$\Rightarrow m = 1$

$P \equiv (a m^{2}, - 2 a m) = (2, - 4)$

$∴$ Shortest distance, PQ = PC – r

$= \sqrt{{(2 - 0)}^{2} + {(- 4 + 6)}^{2}} - 1 = \sqrt{8} - 1 = (2 \sqrt{2} - 1) units$ .

Q 21 :

Let the equation x(x + 2)(12 – k) = 2 have equal roots. Then the distance of the point $(k, \frac{k}{2})$ from the line 3x + 4y + 5 = 0 is [2025]

$5 \sqrt{3}$
$15 \sqrt{5}$
12
15

1

2

3

4

(4)

We have, $(x^{2} + 2 x) (12 - k) = 2$

$\Rightarrow λ x^{2} + 2 λ x - 2 = 0, k \neq 12$ [Let 12 – k = $λ$ ]

For equal roots, D = 0

$\Rightarrow 4 λ^{2} + 8 λ = 0$

$\Rightarrow 4 λ (λ + 2) = 0$

$\Rightarrow λ = 0$ (which is not possible)

or $λ = - 2 \Rightarrow 12 - k = - 2$

$\Rightarrow k = 14$

So, $P (k, \frac{k}{2}) = (14, 7)$

$∴ Distance, d =$ $| \frac{3 \times 14 + 4 \times 7 + 5}{\sqrt{9 + 16}} |$ $= 15$ .

Q 22 :

Consider the lines $x (3 λ + 1) + y (7 λ + 2) = 17 λ + 5$ , $λ$ being a parameter, all passing through a point P. One of these lines (say L) is farthest from the origin. If the distance of L from the point (3, 6) is d, then the value of $d^{2}$ is [2025]

15
10
30
20

1

2

3

4

(4)

We have, $x (3 λ + 1) + y (7 λ + 2) = 17 λ + 5$

$\Rightarrow (x + 2 y - 5) + λ (3 x + 7 y - 17) = 0$

The intersection of family of lines is the point P(1, 2) and let $Q \equiv (3, 6)$ .

Now, $d = P Q = \sqrt{2^{2} + 4^{2}} = \sqrt{20} \Rightarrow d^{2} = 20$

Q 23 :

For an integer $n \geq 2$ , if the arithmetic mean of all coefficient in the binomial expansion of ${(x + y)}^{2 n - 3}$ is 16, then the distance of the point $P (2 n - 1, n^{2} - 4 n)$ from the line x + y = 8 is [2025]

$\sqrt{2}$
$2 \sqrt{2}$
$3 \sqrt{2}$
$5 \sqrt{2}$

1

2

3

4

(3)

Number of terms in ${(x + y)}^{2 n - 3} = 2 n - 3 + 1 = 2 n - 2$

If x = 1, y = 1

Then, sum of all coefficients = $2^{2 n - 3}$

So, arithmetic mean of all coefficients = $\frac{2^{2 n - 3}}{(2 n - 3) + 1}$

$\Rightarrow \frac{2^{2 n - 3}}{2 n - 2} = 16 \Rightarrow 2^{2 n} = 256 (n - 1) \Rightarrow n = 5$

Now, $P (2 n - 1, n^{2} - 4 n) = (9, 5)$

So, distance from P(9, 5) to the line x + y = 8 is

$\frac{(9 + 5 - 8)}{\sqrt{1 + 1}} = \frac{6}{\sqrt{2}} = 3 \sqrt{2} units$ .

Q 24 :

Let ABC be the triangle such that the equation of lines AB and AC be 3y – x = 2 and x + y = 2, respectively, and the points B and C lie on x-axis. If P is the orthocentre of the triangle ABC, then the area of the triangle PBC is equal to [2025]

10
4
8
6

1

2

3

4

(4)

Equation of Altitude AP, which is perpendicular to BC is given by

x = 1 ... (i)

Equation of Altitude BP, which is perpendicular to AC is given by

y – 0 = 1(x + 2) $\Rightarrow$ x – y + 2 = 0 ... (ii)

Hence, $P \equiv (1, 3)$ [From (i) and (ii)]

$∴ Area of △ P B C = \frac{1}{2} \times 4 \times 3 = 6 sq . units$

Q 25 :

If for $θ \in [- \frac{π}{3}, 0]$ , the points $(x, y) = (3 \tan (θ + \frac{π}{3}), 2 \tan (θ + \frac{π}{6}))$ lie on $x y + α x + β y + γ = 0$ , then $α^{2} + β^{2} + γ^{2}$ is equal to [2025]

72
75
80
96

1

2

3

4

(2)

We have, $x = 3 \tan (θ + \frac{π}{3})$

$\Rightarrow x = 3 (\frac{\tan θ + \sqrt{3}}{1 - \sqrt{3} \tan θ})$

$\Rightarrow x - x \sqrt{3} \tan θ = 3 \tan θ + 3 \sqrt{3}$

$\Rightarrow \tan θ = \frac{x - 3 \sqrt{3}}{3 + \sqrt{3} x}$ ... (i)

Also, $y = 2 \tan (θ + \frac{π}{6})$

$\Rightarrow y = 2 \frac{(\tan θ + \frac{1}{\sqrt{3}})}{(1 - \frac{\tan θ}{\sqrt{3}})}$

$\Rightarrow y (\sqrt{3} - \tan θ) = 2 (\sqrt{3} \tan θ + 1)$ ... (ii)

Solving (i) and (ii), we get

$\Rightarrow 2 (\frac{x - 3 \sqrt{3}}{\sqrt{3} + x} + 1) = y (\sqrt{3} - \frac{(x - 3 \sqrt{3})}{\sqrt{3} (\sqrt{3} + x)})$

$\Rightarrow 2 \sqrt{3} (x - 3 \sqrt{3} + x + \sqrt{3}) = y (3 (\sqrt{3} + x) - x + 3 \sqrt{3})$

$\Rightarrow 4 \sqrt{3} x - 12 = y (2 x + 6 \sqrt{3})$

$\Rightarrow x y - 2 \sqrt{3} x + 3 \sqrt{3} y + 6 = 0$

Comparing with $x y + α x + β y + γ$ , we get

$α = - 2 \sqrt{3}, β = 3 \sqrt{3}, γ = 6$

$∴ α^{2} + β^{2} + γ^{2} = 12 + 27 + 36 = 75$ .

Q 26 :

If the orthocenter of the triangle formed by the lines y = x + 1, y = 4x – 8 and y = mx + c is at (3, –1), then m – c is: [2025]

–2
0
4
2

1

2

3

4

(2)

Using lines PQ and QR, we get

Point Q = $(\frac{1 - c}{m - 1}, \frac{1 - c}{m - 1} + 1)$

$m_{Q H} = \frac{\frac{1 - c}{m - 1} + 2}{\frac{1 - c}{m - 1} - 3} = \frac{1 - c + 2 m - 2}{1 - c - 3 m + 3} = - \frac{1}{4} [∵ Q H ⊥ P R]$ ... (i)

$∵ m_{P H} = \frac{5}{0} \to \infty \Rightarrow Slope of line Q R (m) = 0$

Put value of m in equation (i), we get

$\frac{1 - c - 2}{1 - c + 3} = - \frac{1}{4} \Rightarrow c = 0$

$\Rightarrow m - c = 0$ .

Q 27 :

A line passing through the point P(a, 0) makes an acute angle $α$ with the positive x-axis. Let this line be rotated about the point P through an angle $\frac{α}{2}$ in the clock-wise direction. If in the new position, the slope of the line is $2 - \sqrt{3}$ and its distance from the origin is $\frac{1}{\sqrt{2}}$ , then the value of $3 a^{2} \tan^{2} α - 2 \sqrt{3}$ is: [2025]

8
6
5
4

1

2

3

4

(4)

Slope of new line $= 2 - \sqrt{3}$

The new angle with x-axis is $α - \frac{α}{2} = \frac{α}{2}$

$∴ \tan (\frac{α}{2}) = 2 - \sqrt{3} = \tan (\frac{π}{12}) \Rightarrow \frac{α}{2} = \frac{π}{12} \Rightarrow α = \frac{π}{6}$

The new line passes through (a, 0) and has slope $2 - \sqrt{3}$ .

So equation of new line is

$(y - 0) = (2 - \sqrt{3}) (x - a) \Rightarrow y - (2 - \sqrt{3}) x = - (2 - \sqrt{3}) a$

Now, distance of this line from origin is $\frac{1}{\sqrt{2}}$

$\Rightarrow$ $| \frac{0 + (2 - \sqrt{3}) a}{\sqrt{{(2 - \sqrt{3})}^{2} + 1^{2}}} |$ $= \frac{1}{\sqrt{2}} \Rightarrow \frac{(2 - \sqrt{3}) | a |}{\sqrt{8 - 4 \sqrt{3}}} = \frac{1}{\sqrt{2}}$

On squaring both sides, we get $\frac{{(2 - \sqrt{3})}^{2} a^{2}}{8 - 4 \sqrt{3}} = \frac{1}{2}$

$\Rightarrow 2 (7 - 4 \sqrt{3}) a^{2} = 8 - 4 \sqrt{3}$

$\Rightarrow (7 - 4 \sqrt{3}) a^{2} = 4 - 2 \sqrt{3}$

$\Rightarrow a^{2} = \frac{4 - 2 \sqrt{3}}{7 - 4 \sqrt{3}} \times \frac{7 + 4 \sqrt{3}}{7 + 4 \sqrt{3}}$

$= \frac{28 + 16 \sqrt{3} - 14 \sqrt{3} - 24}{49 - 48}$

$a^{2} = 4 + 2 \sqrt{3}$

Now, $3 a^{2} \tan^{2} α - 2 \sqrt{3} = 3 (4 + 2 \sqrt{3}) \tan^{2} \frac{π}{6} - 2 \sqrt{3}$

$= (12 + 6 \sqrt{3}) \cdot \frac{1}{3} - 2 \sqrt{3}$

$= 4 + 2 \sqrt{3} - 2 \sqrt{3} = 4$ .

Q 28 :

Let a be the length of a side of a square OABC with O being the origin. Its side OA makes an acute angle $α$ with the positive x-axis and the equations of its diagonals are $(\sqrt{3} + 1) x + (\sqrt{3} - 1) y = 0$ and $(\sqrt{3} - 1) x - (\sqrt{3} + 1) y + 8 \sqrt{3} = 0$ . Then $a^{2}$ is equal to: [2025]

48
32
24
16

1

2

3

4

(1)

Slope of OB = $\frac{\sqrt{3} + 1}{1 - \sqrt{3}} = \tan 105 °$

Now, $α + 45 ° = 105 ° \Rightarrow α = 60 °$

$∴$ Coordinates of A are (a cos 60°, a sin 60°)

i.e., $(\frac{a}{2}, \frac{\sqrt{3} a}{2})$

Now, A lies on diagonal AC.

$\Rightarrow (\sqrt{3} - 1) \frac{a}{2} - (\sqrt{3} + 1) \frac{\sqrt{3} a}{2} + 8 \sqrt{3} = 0$

$\Rightarrow \frac{a}{2} [\sqrt{3} - 1 - 3 - \sqrt{3}] + 8 \sqrt{3} = 0$

$\Rightarrow \frac{a}{2} [- 4] + 8 \sqrt{3} = 0$

$\Rightarrow a = 4 \sqrt{3}$

$\Rightarrow a^{2} = 48$ .

Q 29 :

Let the triangle PQR be the image of the triangle with vertices (1, 3), (3, 1) and (2, 4) in the line x + 2y = 2. If the centroid of $△$ PQR is the point $(α, β)$ , then $15 (α - β)$ is equal to: [2025]

19
24
21
22

1

2

3

4

(4)

Let P, Q and R be the mirror images of points (1, 3), (3, 1) and (2, 4) in the line x + 2y – 2 = 0.

Coordinates of P is given by

$\frac{x - 1}{1} = \frac{y - 3}{2} = - 2 (\frac{1 + 6 - 2}{5}) i . e ., P (- 1, - 1)$

Similarly, $Q (\frac{9}{5}, \frac{- 7}{5}) and R (\frac{- 6}{5}, \frac{- 12}{5})$

Centroid of $△ P Q R = (\frac{- 2}{15}, - \frac{8}{5})$

$∴ α = \frac{- 2}{15} and β = - \frac{- 8}{5}$

Thus, $15 (α - β) = 15 (\frac{- 2}{15} + \frac{8}{5}) = 22$ .

Q 30 :

A rod of length eight units moves such that its ends A and B always lie on the lines x – y + 2 = 0 and y + 2 = 0, respectively. If the locus of the point P, that divides the rod AB internally in the ratio 2 : 1 is $9 (x^{2} + α y^{2} + β x y + γ x + 28 y) - 76 = 0$ , then $α - β - γ$ is equal to: [2025]

22
21
24
23

1

2

3

4

(4)

Let P(h, k) be the point which divides AB internally in the ratio 2 : 1.

$∴ h = \frac{2 β + α}{3} and k = \frac{- 4 + α + 2}{3}$

$\Rightarrow α = 3 k + 2$

$∴ 2 β = 3 h - α = 3 h - 3 k - 2$

So, AB = 8

$\Rightarrow {(α - β)}^{2} + {(α + 4)}^{2} = 64$

$\Rightarrow (3 k + 2 - {(\frac{3 h - 3 k - 2}{2}))}^{2} + {(3 k + 2 + 4)}^{2} = 64$

$\Rightarrow \frac{{(9 k - 3 h + 6)}^{2}}{4} + {(3 k + 6)}^{2} = 64$

$\Rightarrow 9 [{(3 k - h + 2)}^{2} + 4 {(k + 2)}^{2}] = 64 \times 4$

$\Rightarrow 9 (9 k^{2} + h^{2} + 4 - 6 k h + 12 k - 4 h + 4 k^{2} + 16 + 16 k) = 256$

$\Rightarrow 9 (13 k^{2} + h^{2} - 6 k h + 28 k - 4 h) = 76$

$\Rightarrow 9 (x^{2} + 13 y^{2} - 6 x y - 4 x + 28 y) = 76$

Comparing the equation with

$[9 (x^{2} + α y^{2} + β x y + γ x + 28 y) - 76] = 0$

$\Rightarrow α = 13, β = - 6, γ = - 4$

$∴ α - β - γ = 13 + 6 + 4 = 23$

Topic Question Set

Q 1 :

The vertices of a triangle are A(–1, 3), B(–2, 2) and C(3, –1). A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is [2024]

Q 2 :

Let two straight lines drawn from the origin O intersect the line 3x + 4y = 12 at the points P and Q such that $△ O P Q$ is an isosceles triangle and $∠ P O Q = 90 °$ . If $l = O P^{2} + P Q^{2} + Q O^{2}$ , then the greatest integer less than or equal to $l$ is: [2024]

Q 3 :

Let A(–1, 1) and B(2, 3) be two points and P be a variable point above the line AB such that the area of $△ P A B$ is 10. If the locus of P is ax + by = 15, then 5a + 2b is: [2024]

Q 4 :

The equations of two sides AB and AC of a triangle ABC are 4x + y = 14 and 3x – 2y = 5, respectively. The point $(2, - \frac{4}{3})$ divides the third side BC internally in the ratio 2 : 1, the equation of the side BC is [2024]

Q 5 :

The Portion of the line 4x + 5y = 20 in the first quadrant is trisected by the lines $L_{1}$ and $L_{2}$ passing through the origin. The tangent of an angle between the lines $L_{1}$ and $L_{2}$ is: [2024]

Q 6 :

Let R be the interior region between the lines 3x – y + 1 = 0 and x + 2y – 5 = 0 containing the origin. The set of all values of a, for which the points $(a^{2}, a + 1)$ lie in R, is: [2024]

Q 7 :

In a $△ A B C$ , Suppose y = x is the equation of the bisector of the angle B and the equation of the side AC is 2x – y = 2. If 2AB = BC and the points A and B are respectively (4, 6) and $(α, β)$ , then $α + 2 β$ is equal to [2024]

Q 8 :

The distance of the point (2, 3) from the line 2x – 3y + 28 = 0, measured parallel to the line $\sqrt{3} x - y + 1 = 0$ , is equal to [2024]

Q 9 :

Let A be the point of intersection of the lines 3x + 2y = 14, 5x – y = 6 and B be the point of intersection of the lines 4x + 3y = 8, 6x + y = 5. The distance of the point P(5, –2) from the line AB is [2024]

Q 10 :

A line passing through the point A(9, 0) makes an angle of 30° with the positive direction of x-axis. If this line is rotated about A through an angle of 15° in the clockwise direction, then its equations in the new position is [2024]

Q 11 :

If $x^{2} - y^{2} + 2 h x y + 2 g x + 2 f y + c = 0$ is the locus of a point, which moves such that it is always equidistant from the lines x + 2y + 7 = 0 and 2x – y + 8 = 0, then the value of g + c + h – f equals [2024]

Q 12 :

Let A (a, b), B(3, 4) and C(–6, –8) respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point P(2a + 3, 7b + 5) from the line 2x + 3y – 4 = 0 measured parallel to the line x – 2y – 1 = 0 is [2024]

Q 13 :

Consider a triangle ABC having the vertices A(1, 2), $B (α, β)$ and $C (γ, δ)$ and angles $∠ A B C = \frac{π}{6}$ and $∠ B A C = \frac{2 π}{3}$ . If the points B and C lie on the line y = x + 4, then $α^{2} + γ^{2}$ is equal to __________ [2024]

Q 14 :

Let a ray of light passing through the point (3, 10) reflects on the line 2x + y = 6 and the reflected ray passes through the point (7, 2). If the equation of the incident ray is ax + by + 1 = 0, then $a^{2} + b^{2} + 3 a b$ is equal to __________ . [2024]

Q 15 :

Let ABC be an isosceles triangle in which A is at (–1, 0), $∠ A = \frac{2 π}{3}$ , AB = AC and B is on the positive x-axis. If $B C = 4 \sqrt{3}$ and the line BC intersects the line y = x + 3 at $(α, β)$ , then $\frac{β^{4}}{α^{2}}$ is __________. [2024]

Q 16 :

If the sum of squares of all real values of $α$ , for which the lines 2x – y + 3 = 0, 6x + 3y + 1 = 0 and $α x + 2 y - 2 = 0$ do not form a triangle is p, then the greatest integer less than or equal to p is __________. [2024]

Q 18 :

Let the area of the triangle formed by a straight line L : x + by + c = 0 with co-ordinate axes be 48 square units. If the perpendicular drawn from the origin to the line L makes an angle of 45° with the positive x-axis, then the value of $b^{2} + c^{2}$ is : [2025]

Q 20 :

The shortest distance between the curves $y^{2} = 8 x$ and $x^{2} + y^{2} + 12 y + 35 = 0$ is: [2025]

Q 21 :

Let the equation x(x + 2)(12 – k) = 2 have equal roots. Then the distance of the point $(k, \frac{k}{2})$ from the line 3x + 4y + 5 = 0 is [2025]

Q 22 :

Consider the lines $x (3 λ + 1) + y (7 λ + 2) = 17 λ + 5$ , $λ$ being a parameter, all passing through a point P. One of these lines (say L) is farthest from the origin. If the distance of L from the point (3, 6) is d, then the value of $d^{2}$ is [2025]

Q 23 :

For an integer $n \geq 2$ , if the arithmetic mean of all coefficient in the binomial expansion of ${(x + y)}^{2 n - 3}$ is 16, then the distance of the point $P (2 n - 1, n^{2} - 4 n)$ from the line x + y = 8 is [2025]

Q 24 :

Let ABC be the triangle such that the equation of lines AB and AC be 3y – x = 2 and x + y = 2, respectively, and the points B and C lie on x-axis. If P is the orthocentre of the triangle ABC, then the area of the triangle PBC is equal to [2025]

Q 25 :

If for $θ \in [- \frac{π}{3}, 0]$ , the points $(x, y) = (3 \tan (θ + \frac{π}{3}), 2 \tan (θ + \frac{π}{6}))$ lie on $x y + α x + β y + γ = 0$ , then $α^{2} + β^{2} + γ^{2}$ is equal to [2025]

Q 26 :

If the orthocenter of the triangle formed by the lines y = x + 1, y = 4x – 8 and y = mx + c is at (3, –1), then m – c is: [2025]

Q 29 :

Let the triangle PQR be the image of the triangle with vertices (1, 3), (3, 1) and (2, 4) in the line x + 2y = 2. If the centroid of $△$ PQR is the point $(α, β)$ , then $15 (α - β)$ is equal to: [2025]

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Topic Question Set

Q 1 : The vertices of a triangle are A(–1, 3), B(–2, 2) and C(3, –1). A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is [2024]

Q 2 : Let two straight lines drawn from the origin O intersect the line 3x + 4y = 12 at the points P and Q such that △OPQ is an isosceles triangle and ∠POQ=90°. If l=OP2+PQ2+QO2, then the greatest integer less than or equal to l is: [2024]

Q 3 : Let A(–1, 1) and B(2, 3) be two points and P be a variable point above the line AB such that the area of △PAB is 10. If the locus of P is ax + by = 15, then 5a + 2b is: [2024]

Q 4 : The equations of two sides AB and AC of a triangle ABC are 4x + y = 14 and 3x – 2y = 5, respectively. The point (2,-43) divides the third side BC internally in the ratio 2 : 1, the equation of the side BC is [2024]

Q 5 : The Portion of the line 4x + 5y = 20 in the first quadrant is trisected by the lines L1 and L2 passing through the origin. The tangent of an angle between the lines L1 and L2 is: [2024]

Q 6 : Let R be the interior region between the lines 3x – y + 1 = 0 and x + 2y – 5 = 0 containing the origin. The set of all values of a, for which the points (a2,a+1) lie in R, is: [2024]

Q 7 : In a △ABC, Suppose y = x is the equation of the bisector of the angle B and the equation of the side AC is 2x – y = 2. If 2AB = BC and the points A and B are respectively (4, 6) and (α,β), then α+2β is equal to [2024]

Q 8 : The distance of the point (2, 3) from the line 2x – 3y + 28 = 0, measured parallel to the line 3x-y+1=0, is equal to [2024]

Q 9 : Let A be the point of intersection of the lines 3x + 2y = 14, 5x – y = 6 and B be the point of intersection of the lines 4x + 3y = 8, 6x + y = 5. The distance of the point P(5, –2) from the line AB is [2024]

Q 10 : A line passing through the point A(9, 0) makes an angle of 30° with the positive direction of x-axis. If this line is rotated about A through an angle of 15° in the clockwise direction, then its equations in the new position is [2024]

Q 11 : If x2–y2+2hxy+2gx+2fy+c=0 is the locus of a point, which moves such that it is always equidistant from the lines x + 2y + 7 = 0 and 2x – y + 8 = 0, then the value of g + c + h – f equals [2024]

Q 12 : Let A (a, b), B(3, 4) and C(–6, –8) respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point P(2a + 3, 7b + 5) from the line 2x + 3y – 4 = 0 measured parallel to the line x – 2y – 1 = 0 is [2024]

Q 13 : Consider a triangle ABC having the vertices A(1, 2), B(α,β) and C(γ,δ) and angles ∠ABC=π6 and ∠BAC=2π3. If the points B and C lie on the line y = x + 4, then α2+γ2 is equal to __________ [2024]

Q 14 : Let a ray of light passing through the point (3, 10) reflects on the line 2x + y = 6 and the reflected ray passes through the point (7, 2). If the equation of the incident ray is ax + by + 1 = 0, then a2+b2+3ab is equal to __________ . [2024]

Q 15 : Let ABC be an isosceles triangle in which A is at (–1, 0), ∠A=2π3, AB = AC and B is on the positive x-axis. If BC=43 and the line BC intersects the line y = x + 3 at (α,β), then β4α2 is __________. [2024]

Q 16 : If the sum of squares of all real values of α, for which the lines 2x – y + 3 = 0, 6x + 3y + 1 = 0 and αx+2y–2=0 do not form a triangle is p, then the greatest integer less than or equal to p is __________. [2024]

Q 18 : Let the area of the triangle formed by a straight line L : x + by + c = 0 with co-ordinate axes be 48 square units. If the perpendicular drawn from the origin to the line L makes an angle of 45° with the positive x-axis, then the value of b2+c2 is : [2025]

Q 20 : The shortest distance between the curves y2=8x and x2+y2+12y+35=0 is: [2025]

Q 21 : Let the equation x(x + 2)(12 – k) = 2 have equal roots. Then the distance of the point (k,k2) from the line 3x + 4y + 5 = 0 is [2025]

Q 22 : Consider the lines x(3λ+1)+y(7λ+2)=17λ+5, λ being a parameter, all passing through a point P. One of these lines (say L) is farthest from the origin. If the distance of L from the point (3, 6) is d, then the value of d2 is [2025]

Q 23 : For an integer n≥2, if the arithmetic mean of all coefficient in the binomial expansion of (x+y)2n–3 is 16, then the distance of the point P(2n–1,n2–4n) from the line x + y = 8 is [2025]

Q 24 : Let ABC be the triangle such that the equation of lines AB and AC be 3y – x = 2 and x + y = 2, respectively, and the points B and C lie on x-axis. If P is the orthocentre of the triangle ABC, then the area of the triangle PBC is equal to [2025]

Q 25 : If for θ∈[–π3,0], the points (x,y)=(3tan(θ+π3), 2tan(θ+π6)) lie on xy+αx+βy+γ=0, then α2+β2+γ2 is equal to [2025]

Q 26 : If the orthocenter of the triangle formed by the lines y = x + 1, y = 4x – 8 and y = mx + c is at (3, –1), then m – c is: [2025]

Q 28 : Let a be the length of a side of a square OABC with O being the origin. Its side OA makes an acute angle α with the positive x-axis and the equations of its diagonals are (3+1)x+(3–1)y=0 and (3–1)x–(3+1)y+83=0. Then a2 is equal to: [2025]

Q 29 : Let the triangle PQR be the image of the triangle with vertices (1, 3), (3, 1) and (2, 4) in the line x + 2y = 2. If the centroid of △PQR is the point (α, β), then 15(α–β) is equal to: [2025]

Q 30 : A rod of length eight units moves such that its ends A and B always lie on the lines x – y + 2 = 0 and y + 2 = 0, respectively. If the locus of the point P, that divides the rod AB internally in the ratio 2 : 1 is 9(x2+αy2+βxy+γx+28y)–76=0, then α–β–γ is equal to: [2025]

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Q 1 :

The vertices of a triangle are A(–1, 3), B(–2, 2) and C(3, –1). A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is [2024]

Q 2 :

Let two straight lines drawn from the origin O intersect the line 3x + 4y = 12 at the points P and Q such that $△ O P Q$ is an isosceles triangle and $∠ P O Q = 90 °$ . If $l = O P^{2} + P Q^{2} + Q O^{2}$ , then the greatest integer less than or equal to $l$ is: [2024]

Q 3 :

Let A(–1, 1) and B(2, 3) be two points and P be a variable point above the line AB such that the area of $△ P A B$ is 10. If the locus of P is ax + by = 15, then 5a + 2b is: [2024]

Q 4 :

The equations of two sides AB and AC of a triangle ABC are 4x + y = 14 and 3x – 2y = 5, respectively. The point $(2, - \frac{4}{3})$ divides the third side BC internally in the ratio 2 : 1, the equation of the side BC is [2024]

Q 5 :

The Portion of the line 4x + 5y = 20 in the first quadrant is trisected by the lines $L_{1}$ and $L_{2}$ passing through the origin. The tangent of an angle between the lines $L_{1}$ and $L_{2}$ is: [2024]

Q 6 :

Let R be the interior region between the lines 3x – y + 1 = 0 and x + 2y – 5 = 0 containing the origin. The set of all values of a, for which the points $(a^{2}, a + 1)$ lie in R, is: [2024]

Q 7 :

In a $△ A B C$ , Suppose y = x is the equation of the bisector of the angle B and the equation of the side AC is 2x – y = 2. If 2AB = BC and the points A and B are respectively (4, 6) and $(α, β)$ , then $α + 2 β$ is equal to [2024]

Q 8 :

The distance of the point (2, 3) from the line 2x – 3y + 28 = 0, measured parallel to the line $\sqrt{3} x - y + 1 = 0$ , is equal to [2024]

Q 9 :

Let A be the point of intersection of the lines 3x + 2y = 14, 5x – y = 6 and B be the point of intersection of the lines 4x + 3y = 8, 6x + y = 5. The distance of the point P(5, –2) from the line AB is [2024]

Q 10 :

A line passing through the point A(9, 0) makes an angle of 30° with the positive direction of x-axis. If this line is rotated about A through an angle of 15° in the clockwise direction, then its equations in the new position is [2024]

Q 11 :

If $x^{2} - y^{2} + 2 h x y + 2 g x + 2 f y + c = 0$ is the locus of a point, which moves such that it is always equidistant from the lines x + 2y + 7 = 0 and 2x – y + 8 = 0, then the value of g + c + h – f equals [2024]

Q 12 :

Let A (a, b), B(3, 4) and C(–6, –8) respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point P(2a + 3, 7b + 5) from the line 2x + 3y – 4 = 0 measured parallel to the line x – 2y – 1 = 0 is [2024]

Q 13 :

Consider a triangle ABC having the vertices A(1, 2), $B (α, β)$ and $C (γ, δ)$ and angles $∠ A B C = \frac{π}{6}$ and $∠ B A C = \frac{2 π}{3}$ . If the points B and C lie on the line y = x + 4, then $α^{2} + γ^{2}$ is equal to __________ [2024]

Q 14 :

Let a ray of light passing through the point (3, 10) reflects on the line 2x + y = 6 and the reflected ray passes through the point (7, 2). If the equation of the incident ray is ax + by + 1 = 0, then $a^{2} + b^{2} + 3 a b$ is equal to __________ . [2024]

Q 15 :

Let ABC be an isosceles triangle in which A is at (–1, 0), $∠ A = \frac{2 π}{3}$ , AB = AC and B is on the positive x-axis. If $B C = 4 \sqrt{3}$ and the line BC intersects the line y = x + 3 at $(α, β)$ , then $\frac{β^{4}}{α^{2}}$ is __________. [2024]

Q 16 :

If the sum of squares of all real values of $α$ , for which the lines 2x – y + 3 = 0, 6x + 3y + 1 = 0 and $α x + 2 y - 2 = 0$ do not form a triangle is p, then the greatest integer less than or equal to p is __________. [2024]

Q 18 :

Let the area of the triangle formed by a straight line L : x + by + c = 0 with co-ordinate axes be 48 square units. If the perpendicular drawn from the origin to the line L makes an angle of 45° with the positive x-axis, then the value of $b^{2} + c^{2}$ is : [2025]

Q 20 :

The shortest distance between the curves $y^{2} = 8 x$ and $x^{2} + y^{2} + 12 y + 35 = 0$ is: [2025]

Q 21 :

Let the equation x(x + 2)(12 – k) = 2 have equal roots. Then the distance of the point $(k, \frac{k}{2})$ from the line 3x + 4y + 5 = 0 is [2025]

Q 22 :

Consider the lines $x (3 λ + 1) + y (7 λ + 2) = 17 λ + 5$ , $λ$ being a parameter, all passing through a point P. One of these lines (say L) is farthest from the origin. If the distance of L from the point (3, 6) is d, then the value of $d^{2}$ is [2025]

Q 23 :

For an integer $n \geq 2$ , if the arithmetic mean of all coefficient in the binomial expansion of ${(x + y)}^{2 n - 3}$ is 16, then the distance of the point $P (2 n - 1, n^{2} - 4 n)$ from the line x + y = 8 is [2025]

Q 24 :

Let ABC be the triangle such that the equation of lines AB and AC be 3y – x = 2 and x + y = 2, respectively, and the points B and C lie on x-axis. If P is the orthocentre of the triangle ABC, then the area of the triangle PBC is equal to [2025]

Q 25 :

If for $θ \in [- \frac{π}{3}, 0]$ , the points $(x, y) = (3 \tan (θ + \frac{π}{3}), 2 \tan (θ + \frac{π}{6}))$ lie on $x y + α x + β y + γ = 0$ , then $α^{2} + β^{2} + γ^{2}$ is equal to [2025]

Q 26 :

If the orthocenter of the triangle formed by the lines y = x + 1, y = 4x – 8 and y = mx + c is at (3, –1), then m – c is: [2025]

Q 29 :

Let the triangle PQR be the image of the triangle with vertices (1, 3), (3, 1) and (2, 4) in the line x + 2y = 2. If the centroid of $△$ PQR is the point $(α, β)$ , then $15 (α - β)$ is equal to: [2025]