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

Q 51 :

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

1

2

3

4

(1)

Q 52 :

Let a line L passing through the point P(1,1,1) be perpendicular to the lines $\frac{x - 4}{4} = \frac{y - 1}{1} = \frac{z - 1}{1} and \frac{x - 17}{1} = \frac{y - 71}{1} = \frac{z}{0} .$ Let the line L intersect the $y z$ -plane at the point Q. Another line parallel to L and passing through the point S(1, 0, -1) intersects the $y z$ -plane at the point R. Then the square of the area of the parallelogram PQRS is equal to ________ . [2026]

0%

(6)

Q 53 :

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 54 :

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 55 :

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)

Q 56 :

The sum of all values of $α$ , for which the shortest distance between the lines

$\frac{x + 1}{α} = \frac{y - 2}{- 1} = \frac{z - 4}{- α} and \frac{x}{α} = \frac{y - 1}{2} = \frac{z - 1}{2 α}$ is $\sqrt{2}$ , is [2026]

8
- 6
6
- 8

1

2

3

4

(2)

Q 57 :

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)

Q 58 :

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 51 :

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]

0%

Q 53 :

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 55 :

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 56 :

The sum of all values of $α$ , for which the shortest distance between the lines

$\frac{x + 1}{α} = \frac{y - 2}{- 1} = \frac{z - 4}{- α} and \frac{x}{α} = \frac{y - 1}{2} = \frac{z - 1}{2 α}$ is $\sqrt{2}$ , is [2026]

Q 57 :

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 58 :

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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