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

Q 41 :

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]

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

1

2

3

4

(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 42 :

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]

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

1

2

3

4

(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 43 :

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]

0%

(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 44 :

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]

0%

(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 45 :

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]

0%

(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 46 :

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]

0%

(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 47 :

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]

0%

(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 48 :

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]

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

1

2

3

4

(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 49 :

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

1

2

3

4

(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 50 :

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

1

2

3

4

(3)

$\sin θ = \frac{3}{a}$

$\sin (60 ° + θ) = \frac{9}{a}$

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

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

$a = \sqrt{84}$

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

Topic Question Set