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

Q 11 :

Let the range of the function $f (x) = 6 + 16 \cos x \cdot \cos (\frac{π}{3} - x) \cdot \cos (\frac{π}{3} + x) \cdot \sin 3 x \cdot \cos 6 x, x \in R$ be $[α, β]$ . Then the distance of the point $(α, β)$ from the line 3x + 4y + 12 = 0 is: [2025]

8
11
9
10

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(2)

$f (x) = 6 + 16 (\frac{1}{4} \cos 3 x) \sin 3 x \cdot \cos 6 x$

$[∵ \cos θ \cos (\frac{π}{3} - θ) \cos (\frac{π}{3} + θ) = \frac{1}{4} \cos 3 θ]$

$= 6 + 4 \cos 3 x \sin 3 x \cos 6 x = 6 + \sin 12 x$

$∵$ Range of sin x = [–1, 1]

$∴$ Range of f(x) is [5, 7]

$\Rightarrow (α, β) \equiv (5, 7)$

$∴$ Distance of point (5, 7) from the line 3x + 4y + 12 = 0

$=$ $| \frac{15 + 28 + 12}{5} |$ $= 11$

Q 12 :

Let the lines 3x – 4y – $α$ = 0, 8x –11y – 33 = 0 and 2x – 3y + $λ$ = 0 be concurrent. If the image of the point (1, 2) in the line 2x – 3y + $λ$ = 0 is $(\frac{57}{13}, \frac{- 40}{13})$ , then $| α λ |$ is equal to [2025]

101
113
84
91

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(4)

As the three lines are concurrent,

$| \begin{matrix} 3 & - 4 & - α \\ 8 & - 11 & - 33 \\ 2 & - 3 & λ \end{matrix} |$ $= 0$

$\Rightarrow 3 (- 11 λ - 99) + 4 (8 λ + 66) - α (- 24 + 22) = 0$

$\Rightarrow - 33 λ - 297 + 32 λ + 264 + 2 α = 0$

$\Rightarrow - λ - 33 + 2 α = 0$ ...(i)

As image, (1, 2) w.r.t. 2x – 3y + $λ$ = 0 is $(\frac{57}{13}, \frac{- 40}{13})$

$\frac{\frac{57}{13} - 1}{2} = \frac{\frac{- 40}{13} - 2}{- 3} = - 2 (\frac{2 - 6 + λ}{4 + 9})$

$\Rightarrow \frac{22}{13} = \frac{- 2 (- 4 + λ)}{13} \Rightarrow λ = - 7$

Substitute $λ$ = –7 in (i), we get

$7 - 33 + 2 α = 0 \Rightarrow a = 13$

$∴ | α λ | = | 13 \times (- 7) | = 91$

Q 13 :

Two equal sides of an isosceles triangle are along – x + 2y = 4 and x + y = 4. If m is the slope of its third side, then the sum, of all possible distinct values of m, is: [2025]

$- 2 \sqrt{10}$
12
–6
6

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(4)

Slope of given lines are $m_{1} = \frac{1}{2}$ and $m_{2} = - 1$ .

Since, AB = AC, then $∠$ ABC = $∠$ ACB

Now, angle between AB and BC = angle between AC and BC

$\frac{m - \frac{1}{2}}{1 + \frac{1}{2} (m)} = \frac{- 1 - m}{1 - m}$

$\Rightarrow \frac{2 m - 1}{2 + m} = \frac{m + 1}{m - 1}$

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

$\Rightarrow m^{2} - 6 m - 1 = 0$

Sum of roots = 6

$∴$ Required sum is 6.

Q 14 :

Let the distance between two parallel lines be 5 units and a point P lie between the lines at a unit distance from one of them. An equilateral triangle PQR is formed such that Q lies on one of the parallel lines, while R lies on the other. Then ${(Q R)}^{2}$ is equal to _________. [2025]

(28)

Let $∠ P R N = θ$

$\Rightarrow P R = \cos e c θ and P Q = 4 \sec (30 ° + θ)$

For equilateral $△$ PQR.

Let PR = PQ

$\Rightarrow cosec θ = 4 \sec (θ + 30 °)$

$\Rightarrow \cos (θ + 30 °) = 4 \sin θ$

$\Rightarrow \frac{\sqrt{3}}{2} \cos θ - \frac{1}{2} \sin θ = 4 \sin θ$

$\Rightarrow \tan θ = \frac{1}{3 \sqrt{3}}$

$\Rightarrow cosec θ = \sqrt{28}$

Now, $Q R^{2} = d^{2} = {cosec}^{2} θ = 28$ .

Q 15 :

If $α = 1 + \sum_{r = 1}^{6} {(- 3)}^{r - 1} {}^{12}C_{2 r - 1}$ , then the distance of the point $(12, \sqrt{3})$ from the line $α x - \sqrt{3} y + 1 = 0$ is _________. [2025]

(5)

Given, $α = 1 + \sum_{r = 1}^{6} {(- 3)}^{r - 1} {}^{12}C_{2 r - 1}$

$= 1 + \sum_{r = 1}^{6} {}^{12}C_{2 r - 1} \frac{{(\sqrt{3} i)}^{2 r - 1}}{\sqrt{3} i}$

Let $t = \sqrt{3} i$ , then we have

$α = 1 + \frac{1}{\sqrt{3} i} [{}^{12}C_{1} t + {}^{12}C_{3} t^{3} + . . . + {}^{12}C_{11} t^{11}]$

$= 1 + \frac{1}{\sqrt{3} i} [\frac{{(1 + \sqrt{3} i)}^{12} - {(1 - \sqrt{3} i)}^{12}}{2}]$

$= 1 + \frac{1}{\sqrt{3} i} [\frac{{(- 2 ω^{2})}^{12} - {(2 ω)}^{12}}{2}] = 1$ [ $∵ ω^{3} = 0$ ]

$∴$ Distance of $(12, \sqrt{3})$ from $x - \sqrt{3} y + 1 = 0$ is

$| \frac{12 - \sqrt{3} (\sqrt{3}) + 1}{\sqrt{{(1)}^{2} + {(- \sqrt{3})}^{2}}} |$ $= \frac{12 - 3 + 1}{\sqrt{1 + 3}} = 5$ .

Q 16 :

The straight lines $l_{1}$ and $l_{2}$ pass through the origin and trisect the line segment of the line $L : 9 x + 5 y = 45$ between the axes. If $m_{1}$ and $m_{2}$ are the slopes of the lines $l_{1}$ and $l_{2}$ , then the point of intersection of the line $y = (m_{1} + m_{2}) x$ with L lies on [2023]

$y - x = 5$
$6 x + y = 10$
$y - 2 x = 5$
$6 x - y = 15$

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(1)

Let $l_{1} : y = m_{1} x$ and $l_{2} : y = m_{2} x$ be the two lines which trisects L.

Now, $9 x + 5 y = 45 \Rightarrow \frac{x}{5} + \frac{y}{9} = 1$ ...(i)

Now, L intersects coordinate axes at $A (5, 0)$ and $B (0, 9)$ .

Since $l_{1}$ divides $A B$ in the ratio $2 : 1$

$∴$ $M \equiv (\frac{5}{3}, 6)$

Also, $l_{2}$ divides $A B$ in the ratio $1 : 2$ .

$∴$ $N \equiv (\frac{10}{3}, 3)$

So, $l_{1} : 6 = m_{1} \times \frac{5}{3} \Rightarrow m_{1} = \frac{18}{5}$
and $l_{2} : 3 = m_{2} \times \frac{10}{3} \Rightarrow m_{2} = \frac{9}{10}$

So, equation of line $y = (m_{1} + m_{2}) x$ becomes

$y = (\frac{18}{5} + \frac{9}{10}) x \Rightarrow y = \frac{9}{2} x$ ...(ii)

Now point of intersection of (i) and (ii) is,

$\frac{x}{5} + \frac{x}{2} = 1 \Rightarrow \frac{7 x}{10} = 1 \Rightarrow x = \frac{10}{7}$ $So, y = \frac{45}{7}$

i.e., $(\frac{10}{7}, \frac{45}{7})$ is the point of intersection of (i) & (ii), which satisfies option (1).

Q 17 :

Let $(α, β)$ be the centroid of the triangle formed by the lines $15 x - y = 82, 6 x - 5 y = - 4$ and $9 x + 4 y = 17 .$ Then $α + 2 β$ and $2 α - β$ are the roots of the equation [2023]

$x^{2} - 7 x + 12 = 0$
$x^{2} - 10 x + 25 = 0$
$x^{2} - 14 x + 48 = 0$
$x^{2} - 13 x + 42 = 0$

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(4)

Given lines are

$15 x - y = 82 (i)$

$6 x - 5 y = - 4 (ii)$

$9 x + 4 y = 17 (iii)$

After solving the equations, we get the co-ordinates as $(6, 8), (1, 2)$ and $(5, - 7)$

So, centroid, $(α, β) = (\frac{6 + 1 + 5}{3}, \frac{8 + 2 - 7}{3}) = (4, 1)$

$∴$ $α + 2 β = 4 + 2 = 6; 2 α - β = 7$

So, required equation is $x^{2} - 13 x + 42 = 0$

Q 18 :

If $(α, β)$ is the orthocenter of the triangle ABC with vertices $A (3, - 7)$ , $B (- 1, 2)$ and $C (4, 5)$ , then $9 α - 6 β + 60$ is equal to [2023]

30
35
25
40

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(3)

Draw altitudes BM, CN and AP, then their intersection point is the orthocentre.

Now, let us find the equation of line BM, CN and AP.

Slope of $A C = \frac{12}{1}$

So, slope of $B M = \frac{- 1}{12}$

Equation of BM: $(y - 2) = \frac{- 1}{12} (x + 1)$

$\Rightarrow 12 y - 24 = - x - 1 \Rightarrow 12 y + x = 23$ ...(i)

Now, slope of $A B = \frac{9}{- 4}$

So, slope of $C N = \frac{4}{9}$

Equation of CN: $(y - 5) = \frac{4}{9} (x - 4) \Rightarrow 9 y - 45 = 4 x - 16$

$\Rightarrow 4 x - 9 y - 16 + 45 = 0 \Rightarrow 4 x - 9 y + 29 = 0 ...(ii)$

$Similarly, equation of A P : 5 x + 3 y = - 6 ...(iii)$

$Now, solving (i) and (ii), we get y = \frac{121}{57}, x = \frac{- 141}{57}$

$Also,$ $(\frac{- 141}{57}, \frac{121}{57})$ $satisfies (iii)$

$So, orthocentre, (α, β) = (\frac{- 141}{57}, \frac{121}{57})$

and $9 α - 6 β + 60 = 9 \times (\frac{- 141}{57}) - \frac{6 \times 121}{57} + 60$

$= \frac{- 423}{19} - \frac{242}{19} + \frac{1140}{19} = \frac{475}{19} = 25$

Q 19 :

The combined equation of the two lines $a x + b y + c = 0$ and $a' x + b' y + c' = 0$ can be written as $(a x + b y + c) (a' x + b' y + c') = 0$ . The equation of the angle bisectors of the lines represented by the equation $2 x^{2} + x y - 3 y^{2} = 0$ is [2023]

$3 x^{2} + 5 x y + 2 y^{2} = 0$
$x^{2} - y^{2} - 10 x y = 0$
$3 x^{2} + x y - 2 y^{2} = 0$
$x^{2} - y^{2} + 10 x y = 0$

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(2)

$Given equation of lines is 2 x^{2} + x y - 3 y^{2} = 0$

$Equation of angle bisector of lines a x^{2} + 2 h x y + b y^{2} = 0 is$

$\frac{x^{2} - y^{2}}{x y} = \frac{a - b}{h}$

$Here, a = 2, b = - 3$ and $h = \frac{1}{2}$

$∴$ Equation of angle bisector of given lines is

$\frac{x^{2} - y^{2}}{x y} = \frac{2 + 3}{\frac{1}{2}} = 10$

$\Rightarrow x^{2} - y^{2} = 10 x y i.e., x^{2} - y^{2} - 10 x y = 0$

Q 20 :

A light ray emits from the origin making an angle $30 °$ with the positive $x$ -axis. After getting reflected by the line $x + y = 1,$ if this ray intersects the $x$ -axis at Q, then the abscissa of Q is [2023]

$\frac{2}{(\sqrt{3} - 1)}$
$\frac{\sqrt{3}}{2 (\sqrt{3} + 1)}$
$\frac{2}{3 + \sqrt{3}}$
$\frac{2}{3 - \sqrt{3}}$

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(3)

$Given Slope of incident ray = \tan 30 ° = \frac{1}{\sqrt{3}}$

$So, slope of incident ray = slope of reflected ray = \frac{1}{\sqrt{3}}$

$Now, Equation of line making an angle 30 ° with positive x-axis and passing through origin is,$ $y = \frac{x}{\sqrt{3}}$

$So, line y = \frac{x}{\sqrt{3}} and x + y = 1 intersect at (\frac{\sqrt{3}}{\sqrt{3} + 1}, \frac{1}{\sqrt{3} + 1})$

$So, equation of reflected ray is$

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

$Put y = 0 \Rightarrow x = \frac{2}{3 + \sqrt{3}}$

Q 21 :

Let B and C be the two points on the line $y + x = 0$ such that B and C are symmetric with respect to the origin. Suppose A is a point on $y - 2 x = 2$ such that $∆ A B C$ is an equilateral triangle. Then, the area of $∆ A B C$ is [2023]

$2 \sqrt{3}$
$\frac{8}{\sqrt{3}}$
$3 \sqrt{3}$
$\frac{10}{\sqrt{3}}$

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(2)

$Let point x = α lie on line y - 2 x = 2 and point x = β lie on line y + x = 0$

$Now, slope of A M = \frac{2 + 2 α}{α} = 1$

$\Rightarrow α = - 2$

$A M = 2 \sqrt{2}$

$And β = \frac{2}{\sqrt{3}}$

$A B = 2 \sqrt{2} \times \frac{2}{\sqrt{3}} = \frac{4 \sqrt{2}}{\sqrt{3}}$

$Area of ∆ A B C = \frac{\sqrt{3}}{4} {(A B)}^{2} = \frac{\sqrt{3}}{4} \times \frac{32}{3} = \frac{8}{\sqrt{3}}$

Q 22 :

A straight line cuts off the intercepts OA = $a$ and OB = b on the positive directions of the $x$ -axis and $y$ -axis respectively. If the perpendicular from origin O to this line makes an angle of $\frac{π}{6}$ with the positive direction of the $y$ -axis and the area of $∆ O A B$ is $\frac{98}{3} \sqrt{3},$ then $a^{2} - b^{2}$ is equal to [2023]

$\frac{392}{3}$
$\frac{196}{3}$
$196$
$98$

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(1)

$Let the perpendicular distance be p .$

$The equation of the line A B is given as$

$x \cos \frac{π}{3} + y \sin \frac{π}{3} = p$

$\Rightarrow x \cdot \frac{1}{2} + y \frac{\sqrt{3}}{2} = p$

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

$Intercept form of line A B will be \frac{x}{a} + \frac{y}{b} = 1$ ...(ii)
$Comparing eq (i) and (ii), we get a = 2 p, b = \frac{2 p}{\sqrt{3}}$

$Now, area of triangle O A B = \frac{98}{3} \sqrt{3}$

$S o, \frac{1}{2} a b = \frac{98}{3} \sqrt{3} \Rightarrow \frac{1}{2} \times 2 p \times \frac{2 p}{\sqrt{3}} = \frac{98}{3} \sqrt{3} \Rightarrow 2 p^{2} = 98$

$\Rightarrow p^{2} = 49 \Rightarrow p = 7 (∵ p \geq 0)$

$∴$ $a = 2 \times 7 = 14,$ $b = \frac{2 \times 7}{\sqrt{3}} = \frac{14}{\sqrt{3}}$

$∴$ $a^{2} - b^{2} = {(14)}^{2} - {(\frac{14}{\sqrt{3}})}^{2}$ $= \frac{392}{3}$

Q 23 :

Let the equations of two adjacent sides of a parallelogram ABCD be $2 x - 3 y = - 23$ and $5 x + 4 y = 23$ . If the equation of its one diagonal AC is $3 x + 7 y = 23$ and the distance of A from the other diagonal is $d$ , then $50 d^{2}$ is equal to ________ . [2023]

(529)

$We have, 2 x - 3 y = - 23$ $\dots (i)$

$5 x + 4 y = 23$ $\dots (i i)$

and $3 x + 7 y = 23$ $\dots (i i i)$

$Solve eqn. (i) and (iii), we get A \equiv (- 4, 5)$

$Solve eqn. (ii) and (iii), we get C \equiv (3, 2)$

$Solve eqn. (i) and (ii), we get B \equiv (- 1, 7)$

$Mid point of A C will be (- \frac{1}{2}, \frac{7}{2}) .$

$Equation of diagonal B D is$

$y - \frac{7}{2} = \frac{7 / 2}{- 1 / 2} (x + \frac{1}{2})$

$\Rightarrow 7 x + y = 0$

$Distance of A from diagonal B D is$

$d = \frac{| 7 \times (- 4) + 5 |}{\sqrt{49 + 1}} = \frac{23}{\sqrt{50}}$

$\Rightarrow 50 d^{2} = {(23)}^{2} = 529$

Q 24 :

If the line $l_{1} : 3 y - 2 x = 3$ is the angular bisector of the lines $l_{2} : x - y + 1 = 0 and l_{3} : α x + β y + 17 = 0,$ then $α^{2} + β^{2} - α - β$ is equal to ______ . [2023]

(348)

$Point of intersection of l_{1} : 3 y - 2 x = 3 and l_{2} : x - y + 1 = 0 is P (0, 1),$

$which lies on l_{3} : α x + β y + 17 = 0, β = - 17$

$Consider a random point Q (- 1, 0) on l_{2} : x - y + 1 = 0$

$Image of point Q about l_{1} : 2 x - 3 y + 3 = 0 is Q' (\frac{- 17}{13}, \frac{6}{13}) .$

$which can be calculated by the formula,$

$\frac{x - (- 1)}{2} = \frac{y - 0}{- 3} = \frac{- 2 (- 2 + 3)}{13}$

$Now, Q' lies on l_{3} : α x + β y + 17 = 0$

$∴$ $α = 7$

$Now, α^{2} + β^{2} - α - β = 348$

Q 25 :

The equations of the sides AB, BC and CA of a triangle ABC are: $2 x + y = 0, x + p y = 21 a (a \neq 0)$ and $x - y = 3$ respectively. Let $P (2, a)$ be the centroid of $∆ A B C$ . Then ${(B C)}^{2}$ is equal to _______ . [2023]

(122)

$2 = \frac{1 + α + β + 3}{3}$

$\Rightarrow 6 = 4 + α + β$

$\Rightarrow α + β = 2$

$∴$ $β = 2 - α . . . (i)$

$∴$ $C (5 - α, 2 - α)$

$Now, \frac{- 2 - 2 α + β}{3} = a$

$\Rightarrow - 2 - 2 α + β = 3 a \Rightarrow β - 2 α = 3 a + 2$

$\Rightarrow 2 - α - 2 α = 3 a + 2$

$∴$ $α = - a \Rightarrow β = 2 + a$

$Point B, C lies on x + p y = 21 a$

$So, α - 2 α p = 21 a \Rightarrow - a + 2 a p = 21 a \Rightarrow 2 a p = 22 a$

$\Rightarrow p a = 11 a . . . (i i)$

$∴$ $p = 11, a = 0 (rejected)$

$and β + 3 + p β = 21 a$

$\Rightarrow 5 + a + p (2 + a) = 21 a \Rightarrow 5 + a + 2 p + p a = 21 a$

$\Rightarrow 5 + a + 2 p + 11 a = 21 a \Rightarrow 2 p + 5 = 21 a - 12 a$

$\Rightarrow 2 p + 5 = 9 a \Rightarrow 2 \times 11 + 5 = 9 a$ $∴ a = 3$

$∴$ Point $B$ is $(- 3, 6)$ and $C (8, 5)$

$∴$ $B C = \sqrt{{(8 + 3)}^{2} + {(5 - 6)}^{2}} \Rightarrow {(B C)}^{2} = 122$

Q 26 :

A triangle is formed by the $X$ -axis, $Y$ -axis and the line $3 x + 4 y = 60 .$ Then the number of points $P (a, b)$ which lie strictly inside the triangle, where $a$ is an integer and $b$ is a multiple of $a$ , is ______ . [2023]

(31)

$For point P (a, b) to lie strictly inside the triangle 3 a + 4 b < 60$

Also $a$ is an integer and $b$ is a multiple of $a$

$∴$ $b = n a \Rightarrow 3 a + 4 n a < 60$

$a (3 + 4 n) < 60$

When $a = 1,$ $3 + 4 n < 60,$ $n can take values 1, \dots, 14$

$When a = 2, 2 (3 + 4 n) < 60 \Rightarrow n can take values 1, 2, \dots, 6$ $i.e. 6 point$

$When a = 3, 3 (3 + 4 n) < 60 \Rightarrow n can take values 1, 2, \dots, 4 i.e. 4 point$

$When a = 4, 4 (3 + 4 n) < 60 \Rightarrow n can take values 1, 2 i.e. 2 point$

$When a = 5, 5 (3 + 4 n) < 60 \Rightarrow n can take values 1, 2 i.e. 2 point$

$When a = 6, 6 (3 + 4 n) < 60 \Rightarrow 1 point$

$When a = 7, 7 (3 + 4 n) < 60 \Rightarrow 1 point$

$When a = 8, 8 (3 + 4 n) < 60 \Rightarrow 1 point$

$Total number of points = 14 + 6 + 4 + 2 + 2 + 1 + 1 + 1 = 31$

Q 27 :

A triangle is formed by the tangents at the point (2, 2) on the curves $y^{2} = 2 x and x^{2} + y^{2} = 4 x,$ and the line $x + y + 2 = 0 .$ If $r$ is the radius of its circumcircle, then $r^{2}$ is equal to __________ . [2023]

(10)

$Given curve S_{1} : y^{2} = 2 x and S_{2} : x^{2} + y^{2} = 4 x$

$Point P (2, 2) is common on S_{1} and S_{2}$

$T_{1} is tangent to S_{1} at P$

$T_{1} : 2 y = x + 2$

$\Rightarrow T_{1} : x - 2 y + 2 = 0$

$T_{2} is tangent to S_{2} at P$

$\Rightarrow T_{2} : 2 x + 2 y = 2 (x + 2)$

$\Rightarrow T_{2} : y = 2$

$Line L_{3} : x + y + 2 = 0$

$Now, P Q = a = \sqrt{20}, Q R = b = \sqrt{8}, R P = c = 6$

$Area of ∆ P Q R = Δ = \frac{1}{2} \times 6 \times 2 = 6$

$∴$ $r = \frac{a b c}{4 Δ} = \frac{\sqrt{160}}{4} = \sqrt{10}, r^{2} = 10$

Q 28 :

A rectangle is formed by the lines $x = 0, y = 0, x = 3$ , and $y = 4$ . Let the line $L$ be perpendicular to $3 x + y + 6 = 0$ and divide the area of the rectangle into two equal parts. Then the distance of the point $(\frac{1}{2}, - 5)$ from the line $L$ is equal to: [2026]

$\sqrt{10}$
$2 \sqrt{5}$
$2 \sqrt{10}$
$3 \sqrt{10}$

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(3)

$Line is y = \frac{x}{3} + C$

$Line passes thru (\frac{3}{2}, 2)$

$2 = \frac{1}{2} + C \Rightarrow C = \frac{3}{2}$

$y = \frac{x}{3} + \frac{3}{2}$

$\Rightarrow 6 y = 2 x + 9$

$Line is 2 x - 6 y + 9 = 0$

$Dist =$ $| \frac{1 + 30 + 9}{\sqrt{40}} |$ $= \sqrt{40} = 2 \sqrt{10}$

Q 29 :

Let $(α, β, γ)$ be the co-ordinates of the foot of the perpendicular drawn from the point (5,4,2) on the line $\vec{r} = (- \hat{i} + 3 \hat{j} + \hat{k}) + λ (2 \hat{i} + 3 \hat{j} - \hat{k}) .$ Then the length of the projection of the vector $α \hat{i} + β \hat{j} + γ \hat{k}$ on the vector $6 \hat{i} + 2 \hat{j} + 3 \hat{k}$ is [2026]

18/7
15/7
3
4

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(1)

$\vec{r} = (- \hat{i} + 3 \hat{j} + \hat{k}) + λ (2 \hat{i} + 3 \hat{j} - \hat{k})$

$\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 1}{- 1} = λ$

Any general point $P$ on the line is

$(2 λ - 1, 3 λ + 3, - λ + 1)$

Let the given point is A (5,4,2).

$\bar{A P} (2 λ - 6) \hat{i} + (3 λ - 1) \hat{j} + (- λ - 1) \hat{k}$

$∵$ $\bar{A P}$ $⊥^{r}$ $Line (L)$

$∴$ $\bar{A P} \cdot (2 \hat{i} + 3 \hat{j} - \hat{k}) = 0$

$2 (2 λ - 6) + 3 (3 λ - 1) - 1 (- λ - 1) = 0$

$\Rightarrow λ = 1$

$∴$ $α = 1, β = 6, γ = 0$

Let the vector $\bar{u} = α \hat{i} + β \hat{j} + γ \hat{k}$

$\bar{u} = \hat{i} + 6 \hat{j} + 0 \hat{k}$

$and$ $\bar{w} = 6 \hat{i} + 2 \hat{j} + 3 \hat{k}$

$So projection = \frac{| \bar{u} \cdot \bar{w} |}{| \bar{w} |} = \frac{18}{7}$

Q 30 :

Let a point A lie between the parallel lines $L_{1} and L_{2}$ such that its distances from $L_{1} and L_{2}$ are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle ABC, where the points B and C lie on the lines $L_{1} and L_{2}$ , respectively, is: [2026]

27
$15 \sqrt{6}$
$21 \sqrt{3}$
$12 \sqrt{2}$

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(3)

$\sin θ = \frac{3}{a}$

$\sin (60 ° + θ) = \frac{9}{a}$

$\frac{\sqrt{3}}{2} \cos θ + \frac{1}{2} \sin θ = \frac{9}{a}$

$\sqrt{3} \sqrt{1 - \frac{9}{a^{2}}} + \frac{3}{a} = \frac{18}{a}$

$a = \sqrt{84}$

$Area of ∆ A B C = \frac{\sqrt{3}}{4} a^{2} = \frac{\sqrt{3}}{4} \times 84 = 21 \sqrt{3}$

Q 31 :

Let $A (1, 0), B (2, - 1)$ and $C (\frac{7}{3}, \frac{4}{3})$ be three points. If the equation of the bisector of the angle $A B C$ is $α x + β y = 5,$ then the value of $α^{2} + β^{2}$ is [2026]

10
5
8
13

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4

(1)

$\frac{B D}{D C} = \frac{A B}{A C} = \frac{\sqrt{2} \times 3}{5 \sqrt{2}} = \frac{3}{5}$

$D = (\frac{12}{8}, \frac{4}{8}) = (\frac{3}{2}, \frac{1}{2})$

$Slope of A D = \frac{- 3 / 2}{1 / 2} = - 3$

$3 x + y = 5$

$α = 3, β = 1; α^{2} + β^{2} = 10$

Q 32 :

If the image of the point $P (1, 2, a)$ in the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{7 - z}{2}$ is $Q (5, b, c)$ , then $a^{2} + b^{2} + c^{2}$ is equal to [2026]

298
264
283
293

1

2

3

4

(1)

$Point M \equiv (3, \frac{b}{2} + 1, \frac{c + a}{2}) satisfies the line$

$\frac{3 - 6}{3} = \frac{\frac{b}{2} + 1 - 7}{2} = \frac{\frac{c + a}{2} - 7}{- 2}$

$- 1 = \frac{b - 12}{4} = \frac{c + a - 14}{- 4}$

$\Rightarrow b = 8 . . . (1)$ & $c + a = 18 . . . (2)$

$Now P Q ⊥ L$

$\Rightarrow (4 i + (b - 2) j + (c - a) k) \cdot (3 i + 2 j - 2 k) = 0$

$\Rightarrow 12 + 2 (b - 2) - 2 (c - a) = 0$

$\Rightarrow 6 + (b - 2) - (c - a) = 0$

$\Rightarrow b - c + a + 4 = 0$

$\Rightarrow 8 - c + a + 4 = 0$

$\Rightarrow c + a = 12 . . . (3)$

$From (2) & (3)$

$c = 15, a = 3$

$So a^{2} + b^{2} + c^{2} = 9 + 64 + 225 = 298$

Q 33 :

Let $P (α, β, γ)$ be the point on the line $\frac{x - 1}{2} = \frac{y + 1}{- 3} = z$ at a distance $4 \sqrt{14}$ from the point $(1, - 1, 0)$ and nearer to the origin. Then the shortest distance between the lines $\frac{x - α}{1} = \frac{y - β}{2} = \frac{z - γ}{3} and \frac{x + 5}{2} = \frac{y - 10}{1} = \frac{z - 3}{1}$ , is equal to [2026]

$7 \sqrt{\frac{5}{4}}$
$2 \sqrt{\frac{7}{4}}$
$4 \sqrt{\frac{7}{5}}$
$4 \sqrt{\frac{5}{7}}$

1

2

3

4

(3)

$Let P (2 λ + 1, - 3 λ - 1, λ)$

$Then 4 λ^{2} + 9 λ^{2} + λ^{2} = 16 \cdot 14 \Rightarrow λ = \pm 4$ $\Rightarrow - 4 (nearer to origin)$

$∴$ $P (- 7, 11, - 4)$

$∴$ $Shortest distance = \frac{| \begin{matrix} 2 & - 1 & 7 \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{matrix} |}{| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{matrix} |}$

$= \frac{28}{\sqrt{1 + 25 + 9}} = \frac{4 \sqrt{7}}{\sqrt{5}}$

Q 34 :

Let Q(a,b,c) be the image of the point P(3,2,1) in the line $\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1} .$ Then the distance of Q from the line $\frac{x - 9}{3} = \frac{y - 9}{2} = \frac{z - 5}{- 2}$ is. [2026]

5
7
8
6

1

2

3

4

(2)

$drs of PN = ⟨ r - 2, 2 r - 2, r ⟩$

$1 (r - 2) + 2 (2 r - 2) + 1 (r) = 0$

$6 r = 6 \Rightarrow r = 1$

$∴$ $N \equiv (2, 2, 2)$

$\Rightarrow Q \equiv (1, 2, 3)$

$A Q = \sqrt{64 + 49 + 4} = \sqrt{117}$

$A M = \frac{| 24 + 14 - 4 |}{\sqrt{9 + 4 + 4}} = \frac{34}{\sqrt{17}} = 2 \sqrt{17}$

$∴$ $Q M = \sqrt{117 - 68} = \sqrt{49} = 7$

Q 35 :

Let the angles made with the positive $x$ -axis by two straight lines drawn from the point P(2, 3) and meeting the line $x + y = 6$ at a distance $\sqrt{\frac{2}{3}}$ from the point P be $θ_{1}$ and $θ_{2}$ . Then the value of $(θ_{1} + θ_{2})$ is: [2026]

$\frac{π}{2}$
$\frac{π}{3}$
$\frac{π}{6}$
$\frac{π}{12}$

1

2

3

4

(1)

$Let Q is (\sqrt{\frac{2}{3}} \cos θ + 2, \sqrt{\frac{2}{3}} \sin θ + 3)$

$so, x + y = 6$

$\sqrt{\frac{2}{3}} (\cos θ + \sin θ) + 5 = 6$

$\sin θ + \cos θ = \sqrt{\frac{3}{2}}$

$1 + \sin 2 θ = \frac{3}{2}$

$\sin 2 θ = \frac{1}{2}$

$2 θ = \frac{π}{6}, \frac{5 π}{6}$

$θ = \frac{π}{12}, \frac{5 π}{6}$

$So θ_{1} + θ_{2} = \frac{π}{2}$

Q 36 :

Let the line L pass through the point (−3,5,2) and make equal angles with the positive coordinate axes. If the distance of L from the point (−2,r,1) is $\sqrt{\frac{14}{3}}$ , then the sum of all possible values of r is : [2026]

6
16
12
10

1

2

3

4

(4)

$Equation of line is : \frac{x + 3}{1} = \frac{y - 5}{1} = \frac{z - 2}{1} = λ$

$∴$ $General point R on line is R (λ - 3, λ + 5, λ + 2)$

$\vec{P R} \equiv (λ - 1, λ + 5 - r, λ + 1)$

Now $\vec{P R} \cdot \vec{d} = 0$

$\Rightarrow (λ - 1) 1 + (λ + 5 - r) 1 + (λ + 1) 1 = 0$

$\Rightarrow 3 λ - r + 5 = 0$

$\Rightarrow λ = \frac{r - 5}{3}$

$∴$ $R \equiv (\frac{r - 5}{3} - 3, \frac{r - 5}{3} + 5, \frac{r - 5}{3} + 2)$

$R \equiv (\frac{r - 14}{3}, \frac{r + 10}{3}, \frac{r + 1}{3})$

Now

$P R = \sqrt{\frac{14}{3}} \Rightarrow {(P R)}^{2} = \frac{14}{3}$

$\Rightarrow {(\frac{r - 14}{3} + 2)}^{2} + {(\frac{r + 10}{3} - r)}^{2} + {(\frac{r + 1}{3} - 1)}^{2} = \frac{14}{3}$

$\Rightarrow \frac{{(r - 8)}^{2}}{9} + \frac{{(10 - 2 r)}^{2}}{9} + \frac{{(r - 2)}^{2}}{9} = \frac{14}{3}$

$\Rightarrow (r^{2} - 16 r + 64) + (100 + 4 r^{2} - 40 r) + (r^{2} - 4 r + 4) = 42$

$\Rightarrow 6 r^{2} - 60 r + 126 = 0$

$\Rightarrow r^{2} - 10 r + 21 = 0$

$\Rightarrow r = 3, 7$

$Sum of possible values of r is 10$

Topic Question Set

Q 11 :

Let the range of the function $f (x) = 6 + 16 \cos x \cdot \cos (\frac{π}{3} - x) \cdot \cos (\frac{π}{3} + x) \cdot \sin 3 x \cdot \cos 6 x, x \in R$ be $[α, β]$ . Then the distance of the point $(α, β)$ from the line 3x + 4y + 12 = 0 is: [2025]

Q 12 :

Let the lines 3x – 4y – $α$ = 0, 8x –11y – 33 = 0 and 2x – 3y + $λ$ = 0 be concurrent. If the image of the point (1, 2) in the line 2x – 3y + $λ$ = 0 is $(\frac{57}{13}, \frac{- 40}{13})$ , then $| α λ |$ is equal to [2025]

Q 13 :

Two equal sides of an isosceles triangle are along – x + 2y = 4 and x + y = 4. If m is the slope of its third side, then the sum, of all possible distinct values of m, is: [2025]

Q 14 :

Let the distance between two parallel lines be 5 units and a point P lie between the lines at a unit distance from one of them. An equilateral triangle PQR is formed such that Q lies on one of the parallel lines, while R lies on the other. Then ${(Q R)}^{2}$ is equal to _________. [2025]

Q 15 :

If $α = 1 + \sum_{r = 1}^{6} {(- 3)}^{r - 1} {}^{12}C_{2 r - 1}$ , then the distance of the point $(12, \sqrt{3})$ from the line $α x - \sqrt{3} y + 1 = 0$ is _________. [2025]

Q 17 :

Let $(α, β)$ be the centroid of the triangle formed by the lines $15 x - y = 82, 6 x - 5 y = - 4$ and $9 x + 4 y = 17 .$ Then $α + 2 β$ and $2 α - β$ are the roots of the equation [2023]

Q 18 :

If $(α, β)$ is the orthocenter of the triangle ABC with vertices $A (3, - 7)$ , $B (- 1, 2)$ and $C (4, 5)$ , then $9 α - 6 β + 60$ is equal to [2023]

Q 19 :

The combined equation of the two lines $a x + b y + c = 0$ and $a' x + b' y + c' = 0$ can be written as $(a x + b y + c) (a' x + b' y + c') = 0$ . The equation of the angle bisectors of the lines represented by the equation $2 x^{2} + x y - 3 y^{2} = 0$ is [2023]

Q 20 :

A light ray emits from the origin making an angle $30 °$ with the positive $x$ -axis. After getting reflected by the line $x + y = 1,$ if this ray intersects the $x$ -axis at Q, then the abscissa of Q is [2023]

Q 21 :

Let B and C be the two points on the line $y + x = 0$ such that B and C are symmetric with respect to the origin. Suppose A is a point on $y - 2 x = 2$ such that $∆ A B C$ is an equilateral triangle. Then, the area of $∆ A B C$ is [2023]

Q 23 :

Let the equations of two adjacent sides of a parallelogram ABCD be $2 x - 3 y = - 23$ and $5 x + 4 y = 23$ . If the equation of its one diagonal AC is $3 x + 7 y = 23$ and the distance of A from the other diagonal is $d$ , then $50 d^{2}$ is equal to ________ . [2023]

Q 24 :

If the line $l_{1} : 3 y - 2 x = 3$ is the angular bisector of the lines $l_{2} : x - y + 1 = 0 and l_{3} : α x + β y + 17 = 0,$ then $α^{2} + β^{2} - α - β$ is equal to ______ . [2023]

Q 25 :

The equations of the sides AB, BC and CA of a triangle ABC are: $2 x + y = 0, x + p y = 21 a (a \neq 0)$ and $x - y = 3$ respectively. Let $P (2, a)$ be the centroid of $∆ A B C$ . Then ${(B C)}^{2}$ is equal to _______ . [2023]

Q 26 :

A triangle is formed by the $X$ -axis, $Y$ -axis and the line $3 x + 4 y = 60 .$ Then the number of points $P (a, b)$ which lie strictly inside the triangle, where $a$ is an integer and $b$ is a multiple of $a$ , is ______ . [2023]

Q 27 :

A triangle is formed by the tangents at the point (2, 2) on the curves $y^{2} = 2 x and x^{2} + y^{2} = 4 x,$ and the line $x + y + 2 = 0 .$ If $r$ is the radius of its circumcircle, then $r^{2}$ is equal to __________ . [2023]

Q 28 :

A rectangle is formed by the lines $x = 0, y = 0, x = 3$ , and $y = 4$ . Let the line $L$ be perpendicular to $3 x + y + 6 = 0$ and divide the area of the rectangle into two equal parts. Then the distance of the point $(\frac{1}{2}, - 5)$ from the line $L$ is equal to: [2026]

Q 30 :

Let a point A lie between the parallel lines $L_{1} and L_{2}$ such that its distances from $L_{1} and L_{2}$ are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle ABC, where the points B and C lie on the lines $L_{1} and L_{2}$ , respectively, is: [2026]

Q 31 :

Let $A (1, 0), B (2, - 1)$ and $C (\frac{7}{3}, \frac{4}{3})$ be three points. If the equation of the bisector of the angle $A B C$ is $α x + β y = 5,$ then the value of $α^{2} + β^{2}$ is [2026]

Q 32 :

If the image of the point $P (1, 2, a)$ in the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{7 - z}{2}$ is $Q (5, b, c)$ , then $a^{2} + b^{2} + c^{2}$ is equal to [2026]

Q 34 :

Let Q(a,b,c) be the image of the point P(3,2,1) in the line $\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1} .$ Then the distance of Q from the line $\frac{x - 9}{3} = \frac{y - 9}{2} = \frac{z - 5}{- 2}$ is. [2026]

Q 35 :

Let the angles made with the positive $x$ -axis by two straight lines drawn from the point P(2, 3) and meeting the line $x + y = 6$ at a distance $\sqrt{\frac{2}{3}}$ from the point P be $θ_{1}$ and $θ_{2}$ . Then the value of $(θ_{1} + θ_{2})$ is: [2026]

Q 36 :

Let the line L pass through the point (−3,5,2) and make equal angles with the positive coordinate axes. If the distance of L from the point (−2,r,1) is $\sqrt{\frac{14}{3}}$ , then the sum of all possible values of r is : [2026]

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Topic Question Set

Q 11 : Let the range of the function f(x)=6+16 cos x·cos(π3–x)·cos(π3+x)·sin 3x·cos 6x, x∈R be [α, β]. Then the distance of the point (α, β) from the line 3x + 4y + 12 = 0 is: [2025]

Q 12 : Let the lines 3x – 4y – α= 0, 8x –11y – 33 = 0 and 2x – 3y + λ = 0 be concurrent. If the image of the point (1, 2) in the line 2x – 3y + λ = 0 is (5713,–4013), then |αλ| is equal to [2025]

Q 13 : Two equal sides of an isosceles triangle are along – x + 2y = 4 and x + y = 4. If m is the slope of its third side, then the sum, of all possible distinct values of m, is: [2025]

Q 14 : Let the distance between two parallel lines be 5 units and a point P lie between the lines at a unit distance from one of them. An equilateral triangle PQR is formed such that Q lies on one of the parallel lines, while R lies on the other. Then (QR)2 is equal to _________. [2025]

Q 15 : If α=1+∑r=16(–3)r–1 C2r–112, then the distance of the point (12,3) from the line αx–3y+1=0 is _________. [2025]

Q 16 : The straight lines l1 and l2 pass through the origin and trisect the line segment of the line L: 9x+5y=45 between the axes. If m1 and m2 are the slopes of the lines l1 and l2, then the point of intersection of the line y=(m1+m2)x with L lies on [2023]

Q 17 : Let (α,β) be the centroid of the triangle formed by the lines 15x-y=82,6x-5y=-4 and 9x+4y=17. Then α+2β and 2α-β are the roots of the equation [2023]

Q 18 : If (α,β) is the orthocenter of the triangle ABC with vertices A(3,-7), B(-1,2) and C(4,5), then 9α-6β+60 is equal to [2023]

Q 19 : The combined equation of the two lines ax+by+c=0 and a'x+b'y+c'=0 can be written as (ax+by+c)(a'x+b'y+c')=0. The equation of the angle bisectors of the lines represented by the equation 2x2+xy-3y2=0 is [2023]

Q 20 : A light ray emits from the origin making an angle 30° with the positive x-axis. After getting reflected by the line x+y=1, if this ray intersects the x-axis at Q, then the abscissa of Q is [2023]

Q 21 : Let B and C be the two points on the line y+x=0 such that B and C are symmetric with respect to the origin. Suppose A is a point on y-2x=2 such that ∆ABC is an equilateral triangle. Then, the area of ∆ABC is [2023]

Q 23 : Let the equations of two adjacent sides of a parallelogram ABCD be 2x-3y=-23 and 5x+4y=23. If the equation of its one diagonal AC is 3x+7y=23 and the distance of A from the other diagonal is d, then 50d2 is equal to ________ . [2023]

Q 24 : If the line l1:3y-2x=3 is the angular bisector of the lines l2:x-y+1=0 and l3:αx+βy+17=0, then α2+β2-α-β is equal to ______ . [2023]

Q 25 : The equations of the sides AB, BC and CA of a triangle ABC are: 2x+y=0, x+py=21a(a≠0) and x-y=3 respectively. Let P(2,a) be the centroid of ∆ABC. Then (BC)2 is equal to _______ . [2023]

Q 26 : A triangle is formed by the X-axis, Y-axis and the line 3x+4y=60. Then the number of points P(a,b) which lie strictly inside the triangle, where a is an integer and b is a multiple of a, is ______ . [2023]

Q 27 : A triangle is formed by the tangents at the point (2, 2) on the curves y2=2x and x2+y2=4x, and the line x+y+2=0. If r is the radius of its circumcircle, then r2 is equal to __________ . [2023]

Q 28 : A rectangle is formed by the lines x=0, y=0, x=3, and y=4. Let the line L be perpendicular to 3x+y+6=0 and divide the area of the rectangle into two equal parts. Then the distance of the point (12,-5) from the line L is equal to: [2026]

Q 29 : Let (α,β,γ) be the co-ordinates of the foot of the perpendicular drawn from the point (5,4,2) on the line r→=(-i^+3j^+k^)+λ(2i^+3j^-k^). Then the length of the projection of the vector αi^+βj^+γk^ on the vector 6i^+2j^+3k^ is [2026]

Q 30 : Let a point A lie between the parallel lines L1 and L2 such that its distances from L1 and L2 are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle ABC, where the points B and C lie on the lines L1 and L2, respectively, is: [2026]

Q 31 : Let A(1,0), B(2,-1) and C(73,43) be three points. If the equation of the bisector of the angle ABC is αx+βy=5, then the value of α2+β2 is [2026]

Q 32 : If the image of the point P(1,2,a) in the line x-63=y-72=7-z2 is Q(5,b,c), then a2+b2+c2 is equal to [2026]

Q 33 : Let P(α,β,γ) be the point on the line x-12=y+1-3=z at a distance 414 from the point (1,-1,0) and nearer to the origin. Then the shortest distance between the lines x-α1=y-β2=z-γ3 and x+52=y-101=z-31, is equal to [2026]

Q 34 : Let Q(a,b,c) be the image of the point P(3,2,1) in the line x-11=y2=z-11. Then the distance of Q from the line x-93=y-92=z-5-2 is. [2026]

Q 35 : Let the angles made with the positive x-axis by two straight lines drawn from the point P(2, 3) and meeting the line x+y=6 at a distance 23 from the point P be θ1 and θ2. Then the value of (θ1+θ2) is: [2026]

Q 36 : Let the line L pass through the point (−3,5,2) and make equal angles with the positive coordinate axes. If the distance of L from the point (−2,r,1) is 143, then the sum of all possible values of r is : [2026]

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Q 11 :

Let the range of the function $f (x) = 6 + 16 \cos x \cdot \cos (\frac{π}{3} - x) \cdot \cos (\frac{π}{3} + x) \cdot \sin 3 x \cdot \cos 6 x, x \in R$ be $[α, β]$ . Then the distance of the point $(α, β)$ from the line 3x + 4y + 12 = 0 is: [2025]

Q 12 :

Let the lines 3x – 4y – $α$ = 0, 8x –11y – 33 = 0 and 2x – 3y + $λ$ = 0 be concurrent. If the image of the point (1, 2) in the line 2x – 3y + $λ$ = 0 is $(\frac{57}{13}, \frac{- 40}{13})$ , then $| α λ |$ is equal to [2025]

Q 13 :

Two equal sides of an isosceles triangle are along – x + 2y = 4 and x + y = 4. If m is the slope of its third side, then the sum, of all possible distinct values of m, is: [2025]

Q 14 :

Let the distance between two parallel lines be 5 units and a point P lie between the lines at a unit distance from one of them. An equilateral triangle PQR is formed such that Q lies on one of the parallel lines, while R lies on the other. Then ${(Q R)}^{2}$ is equal to _________. [2025]

Q 15 :

If $α = 1 + \sum_{r = 1}^{6} {(- 3)}^{r - 1} {}^{12}C_{2 r - 1}$ , then the distance of the point $(12, \sqrt{3})$ from the line $α x - \sqrt{3} y + 1 = 0$ is _________. [2025]

Q 17 :

Let $(α, β)$ be the centroid of the triangle formed by the lines $15 x - y = 82, 6 x - 5 y = - 4$ and $9 x + 4 y = 17 .$ Then $α + 2 β$ and $2 α - β$ are the roots of the equation [2023]

Q 18 :

If $(α, β)$ is the orthocenter of the triangle ABC with vertices $A (3, - 7)$ , $B (- 1, 2)$ and $C (4, 5)$ , then $9 α - 6 β + 60$ is equal to [2023]

Q 19 :

The combined equation of the two lines $a x + b y + c = 0$ and $a' x + b' y + c' = 0$ can be written as $(a x + b y + c) (a' x + b' y + c') = 0$ . The equation of the angle bisectors of the lines represented by the equation $2 x^{2} + x y - 3 y^{2} = 0$ is [2023]

Q 20 :

A light ray emits from the origin making an angle $30 °$ with the positive $x$ -axis. After getting reflected by the line $x + y = 1,$ if this ray intersects the $x$ -axis at Q, then the abscissa of Q is [2023]

Q 21 :

Let B and C be the two points on the line $y + x = 0$ such that B and C are symmetric with respect to the origin. Suppose A is a point on $y - 2 x = 2$ such that $∆ A B C$ is an equilateral triangle. Then, the area of $∆ A B C$ is [2023]

Q 23 :

Let the equations of two adjacent sides of a parallelogram ABCD be $2 x - 3 y = - 23$ and $5 x + 4 y = 23$ . If the equation of its one diagonal AC is $3 x + 7 y = 23$ and the distance of A from the other diagonal is $d$ , then $50 d^{2}$ is equal to ________ . [2023]

Q 24 :

If the line $l_{1} : 3 y - 2 x = 3$ is the angular bisector of the lines $l_{2} : x - y + 1 = 0 and l_{3} : α x + β y + 17 = 0,$ then $α^{2} + β^{2} - α - β$ is equal to ______ . [2023]

Q 25 :

The equations of the sides AB, BC and CA of a triangle ABC are: $2 x + y = 0, x + p y = 21 a (a \neq 0)$ and $x - y = 3$ respectively. Let $P (2, a)$ be the centroid of $∆ A B C$ . Then ${(B C)}^{2}$ is equal to _______ . [2023]

Q 26 :

A triangle is formed by the $X$ -axis, $Y$ -axis and the line $3 x + 4 y = 60 .$ Then the number of points $P (a, b)$ which lie strictly inside the triangle, where $a$ is an integer and $b$ is a multiple of $a$ , is ______ . [2023]

Q 27 :

A triangle is formed by the tangents at the point (2, 2) on the curves $y^{2} = 2 x and x^{2} + y^{2} = 4 x,$ and the line $x + y + 2 = 0 .$ If $r$ is the radius of its circumcircle, then $r^{2}$ is equal to __________ . [2023]

Q 28 :

A rectangle is formed by the lines $x = 0, y = 0, x = 3$ , and $y = 4$ . Let the line $L$ be perpendicular to $3 x + y + 6 = 0$ and divide the area of the rectangle into two equal parts. Then the distance of the point $(\frac{1}{2}, - 5)$ from the line $L$ is equal to: [2026]

Q 30 :

Let a point A lie between the parallel lines $L_{1} and L_{2}$ such that its distances from $L_{1} and L_{2}$ are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle ABC, where the points B and C lie on the lines $L_{1} and L_{2}$ , respectively, is: [2026]

Q 31 :

Let $A (1, 0), B (2, - 1)$ and $C (\frac{7}{3}, \frac{4}{3})$ be three points. If the equation of the bisector of the angle $A B C$ is $α x + β y = 5,$ then the value of $α^{2} + β^{2}$ is [2026]

Q 32 :

If the image of the point $P (1, 2, a)$ in the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{7 - z}{2}$ is $Q (5, b, c)$ , then $a^{2} + b^{2} + c^{2}$ is equal to [2026]

Q 34 :

Let Q(a,b,c) be the image of the point P(3,2,1) in the line $\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1} .$ Then the distance of Q from the line $\frac{x - 9}{3} = \frac{y - 9}{2} = \frac{z - 5}{- 2}$ is. [2026]

Q 35 :

Let the angles made with the positive $x$ -axis by two straight lines drawn from the point P(2, 3) and meeting the line $x + y = 6$ at a distance $\sqrt{\frac{2}{3}}$ from the point P be $θ_{1}$ and $θ_{2}$ . Then the value of $(θ_{1} + θ_{2})$ is: [2026]

Q 36 :

Let the line L pass through the point (−3,5,2) and make equal angles with the positive coordinate axes. If the distance of L from the point (−2,r,1) is $\sqrt{\frac{14}{3}}$ , then the sum of all possible values of r is : [2026]