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

Q 31 :

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

1

2

3

4

(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 32 :

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

1

2

3

4

(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 33 :

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

1

2

3

4

(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 34 :

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]

0%

(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 35 :

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]

0%

(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 36 :

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]

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

1

2

3

4

(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 37 :

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]

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

1

2

3

4

(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 38 :

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

1

2

3

4

(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 39 :

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]

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

1

2

3

4

(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 40 :

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]

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

1

2

3

4

(3)

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

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

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

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

$So, equation of reflected ray is$

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

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

Topic Question Set

Q 31 :

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 32 :

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 33 :

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 34 :

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]

0%

Q 35 :

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]

0%

Q 37 :

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 38 :

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 39 :

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 40 :

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]

Want Us to Email you About Special Offers & Updates?