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

Q 1 :

Let a variable line of slope m > 0 passing through the point (4, –9) intersect the coordinate axes at the points A and B. The minimum value of the sum of the distances of A and B from the origin is [2024]

10
25
30
15

1

2

3

4

(2)

Equation of line passing through (4, –9) and having slope m is given by

(y + 9) = m(x – 4)

Since it intersect the coordinate axes at points A and B so at A, y = 0

$∴ A = (\frac{9 + 4 m}{m}, 0)$

Similarly at B, x = 0

$∴ B = (0, - 4 m - 9)$

Now, let O(0, 0) be origin, then

$O A + O B = \sqrt{{(\frac{9 + 4 m}{m})}^{2}} + \sqrt{{(9 + 4 m)}^{2}}$

$= \frac{9 + 4 m}{m} + 9 + 4 m$ $[∵ m > 0]$

$= 13 + \frac{9}{m} + 4 m$

Now, $\frac{\frac{9}{m} + 4 m}{2} \geq \sqrt{\frac{9}{m} \cdot 4 m}$ $[∵ A . M . \geq G . M .]$

$\Rightarrow \frac{9}{m} + 4 m \geq 12$

$∴ O A + O B \geq 13 + 12 i . e . 25$

Q 2 :

If the locus of the point, whose distance from the point (2, 1) and (1, 3) are in the ratio 5 : 4, is $a x^{2} + b y^{2} + c x y + d x + e y + 170 = 0$ , then the value of $a^{2} + 2 b + 3 c + 4 d + e$ is equal to : [2024]

5
– 27
37
437

1

2

3

4

(3)

Let given point be (h, k) whose distance from (2, 1) and (1, 3) is in ratio 5 : 4.

$\Rightarrow \frac{{(h - 2)}^{2} + {(k - 1)}^{2}}{{(h - 1)}^{2} + {(k - 3)}^{2}} = \frac{25}{16}$

$\Rightarrow \frac{h^{2} + k^{2} - 4 h - 2 k + 5}{h^{2} + k^{2} - 2 h - 6 k + 10} = \frac{25}{16}$

On replacing h by x and k by y, we get

$16 x^{2} + 16 y^{2} - 64 x - 32 y + 80 = 25 x^{2} + 25 y^{2} - 50 x - 150 y + 250$

$\Rightarrow 9 x^{2} + 9 y^{2} + 14 x - 118 y + 170 = 0$

On comparing, we get

$a = 9, b = 9, c = 0, d = 14, e = - 118$

$∴ a^{2} + 2 b + 3 c + 4 d + e = 81 + 18 + 56 - 118 = 37$

Q 3 :

If the line segment joining the points (5, 2) and (2, a) subtends an angle $\frac{π}{4}$ at the origin, then the absolute value of the product of all possible values of a is: [2024]

8
2
4
6

1

2

3

4

(3)

$m_{1} = \frac{2}{5}, m_{2} = \frac{a}{2}$

Angle between the lines is $π / 4$

Since, $\tan θ =$ $| \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} |$

$\Rightarrow$ $\tan \frac{π}{4} =$ $| \frac{\frac{2}{5} - \frac{a}{2}}{1 + \frac{2}{5} \times \frac{a}{2}} |$ $\Rightarrow$ $| \frac{\frac{2}{5} - \frac{a}{2}}{1 + \frac{a}{5}} |$ $= 1$

$\Rightarrow$ $| \frac{4 - 5 a}{10 + 2 a} |$ $= 1 \Rightarrow \frac{4 - 5 a}{10 + 2 a} = \pm 1$

$\Rightarrow \frac{4 - 5 a}{10 + 2 a} = 1 o r \frac{4 - 5 a}{10 + 2 a} = - 1$

$\Rightarrow 4 - 5 a = 10 + 2 a or 4 - 5 a = - 10 - 2 a$

$\Rightarrow a = - \frac{6}{7} or a = \frac{14}{3}$

$∴$ Required product $=$ $| - \frac{6}{7} \times \frac{14}{3} |$ $= 4$ .

Q 4 :

A ray of light coming from the point P(1, 2) gets reflected from the point Q on the x-axis and then passes through the point R(4, 3). If the point S(h, k) is such that PQRS is a parallelogram, then $h k^{2}$ is equal to: [2024]

70
90
60
80

1

2

3

4

(1)

Slope of $P' R = \frac{3 - (- 2)}{4 - 1} = \frac{5}{3}$

Equation of $P' R : y + 2 = \frac{5}{3} (x - 1)$

Since, point Q is on x-axis.

Let $Q \equiv (a, 0)$

Also, point Q lies P'R

$\Rightarrow \frac{6}{5} = a - 1 \Rightarrow a = \frac{11}{5}$

Now, PQRS is a parallelogram

$∴ \frac{h + a}{2} = \frac{4 + 1}{2} \Rightarrow h = 5 - \frac{11}{5} = \frac{14}{5}$

and $\frac{2 + 3}{2} = \frac{k}{2} \Rightarrow k = 5$

$∴ h k^{2} = \frac{14}{5} \times 25 = 70$

Q 5 :

Two vertices of a triangle ABC are A(3, –1) and B(–2, 3) and its orthocentre is P(1, 1). If the coordinates of the point C are $(α, β)$ and the centre of the circle circumscribing the triangle PAB is (h, k), then the value of $(α + β) + 2 (h + k)$ equals [2024]

15
5
51
81

1

2

3

4

(2)

Slope of $A B = \frac{- 1 - 3}{3 + 2} = \frac{- 4}{5}$

Slope of $P E = \frac{5}{4}$

Equation of PC is given $y - 1 = \frac{5}{4} (x - 1)$

$\Rightarrow$ 4y – 4 = 5x – 5

$\Rightarrow$ 5x – 4y – 1 = 0 ... (i)

Now slope of $A P = \frac{- 1 - 1}{3 - 1} = - 1$

$\Rightarrow$ Slope of BC = 1

Equation of BC is given by

y – 3 = 1(x + 2)

$\Rightarrow$ x – y + 5 = 0 ... (ii)

Solving (i) and (ii), we get

$β = y = 26 [∵ C is point of intersection of P C and B C]$

and $α = x = 21$

Centre of circle circumseribing $∆ P A B$ is point of intersection of perpendicular bisector of AP, AB and AC.

Equation of perpendicular bisector of AP

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

$\Rightarrow$ y – 0 = x – 2

$\Rightarrow$ y = x – 2 ... (iii)

Equation of perpendicular bisector of AB

$y - (\frac{3 - 1}{2}) = \frac{5}{4} (x - (\frac{- 2 + 3}{2}))$

$\Rightarrow y - 1 = \frac{5}{4} (x - \frac{1}{2})$ ... (iv)

On solving (iii) and (iv), we get

$x = h = \frac{- 19}{2} and y = k = \frac{- 23}{2}$

$\Rightarrow 2 (h + k) = - 42$

So, $(α + β) + 2 (h + k) = 47 - 42 = 5$ .

Q 6 :

Let $α, β, γ, δ \in Z$ and let $A (α, β), B (1, 0), C (γ, δ)$ and D(1, 2) be the vertices of a parallelogram ABCD. If $A B = \sqrt{10}$ and the points A and C lie on the line 3y = 2x + 1, then $2 (α + β + γ + δ)$ is equal to [2024]

12
10
8
5

1

2

3

4

(3)

Givem, $A (α, β), B (1, 0), C (γ, δ)$ and D(1, 2).

Using mid point Formula.

$α + γ = 2$

$β + δ = 2$

So, $α + β + γ + δ = 2 + 2$

$α + β + γ + δ = 4$

$∴ 2 (α + β + γ + δ) = 2 (4) = 8$ .

Q 7 :

Let $A (- 2, - 1), B (1, 0), C (α, β)$ and $D (γ, δ)$ be the vertices of a parallelogram ABCD. If the point C lies on 2x – y = 5 and the point D lies on 3x – 2y = 6, then the value of $| α + β + γ + δ |$ is equal to __________. [2024]

(32)

We have, A(–2, –1), B(1, 0), $C (α, β)$ and $D (γ, δ)$ be the vertices of a parallelogram ABCD.

Since, $C (α, β)$ lies on 2x – y = 5

$∴ 2 α - β = 5$ ... (i)

Also, $D (γ, δ)$ lies on the line 3x – 2y = 6

So, $3 γ - 2 δ = 6$ ... (ii)

Now, $\frac{γ + 1}{2} = \frac{α - 2}{2} and \frac{δ + 0}{2} = \frac{β - 1}{2}$

$γ + 1 = α - 2 and δ = β - 1$

$γ = α - 3$ ... (iii)

From (ii), $3 (α - 3) - 2 (β - 1) = 6$

Solving (i) and (iii), we get

$α = - 3; β = - 11; γ = - 6; δ = - 12$

$∴ | α + β + γ + δ | = | - 3 - 11 - 6 - 12 | = 32$ .

Q 8 :

Let the three sides of a triangle are on the lines 4x – 7y + 10 = 0, x + y = 5 and 7x + 4y = 15. Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines x = 0, y = 0 and x + y = 1 is [2025]

$\sqrt{5}$
20
5
$\sqrt{20}$

1

2

3

4

(1)

Since the lines 4x –7y + 10 = 0 and 7x + 4y = 15 are perpendicular to each other.

So, the triangle formed by lines 4x –7y + 10 = 0, 7x + 4y = 15 and x + y = 5 is right angled triangle.

So, orthrocentre = Point of intersection of 4x – 7y + 10 = 0 and 7x + 4y = 15 which is, B = (1, 2)

Also, triangle formed by sides x = 0, y = 0, x + y = 1 is right angled triangle.

So, orthrocentre, Q = (0, 0)

$∴$ Distance between Q and $B = \sqrt{1 + 4} = \sqrt{5} units .$ .

Q 9 :

Let the area of a $△$ PQR with vertices P(5, 4), Q(–2, 4) and R(a, b) be 35 square units. If its orthocenter and centroid are $O (2, \frac{14}{5})$ and C(c, d) respectively, then c + 2d is equal to [2025]

2
3
$\frac{7}{3}$
$\frac{8}{3}$

1

2

3

4

(2)

Since, $P R ⊥ Q T$

$∴ \frac{b - 4}{a - 5} \times \frac{- 6 / 5}{4} = - 1 \Rightarrow \frac{b - 4}{a - 5} \times \frac{3}{10} = 1$

$\Rightarrow 10 a - 3 b = 38$ ... (i)

Also, $P S ⊥ Q R$

$∴ \frac{- 6 / 5}{- 3} \times \frac{b - 4}{a + 2} = - 1$

$\Rightarrow 5 a + 2 b = - 2$ ... (ii)

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

a = 2, b = - 6

Now, centroid of $△$ PQR = (c, d)

$\Rightarrow (\frac{5 - 2 + 2}{3}, \frac{4 + 4 - 6}{3}) = (c, d)$

$\Rightarrow c = \frac{5}{3}, d = \frac{2}{3}$

$∴ c + 2 d = \frac{5}{3} + \frac{4}{3} = 3$ .

Q 10 :

Let the points $(\frac{11}{2}, α)$ lie on or inside the triangle with sides x + y = 11, x + 2y = 16 and 2x + 3y = 29. Then the product of the smallest and the largest values of $α$ is equal to : [2025]

22
55
33
44

1

2

3

4

(3)

Given that a triangle bounded by lines $L_{1}$ : x + y = 11, $L_{2}$ : x + 2y =16 and $L_{3}$ : 2x + 3y = 29.

The region is given by

Using $L_{1}$ : When $x = \frac{11}{2}$ , y = 11 – x

$\Rightarrow α = 11 - \frac{11}{2} = \frac{11}{2}$ (minimum)

Using $L_{3}$ : when x = 11/2, 3y = 29 – 2x

$\Rightarrow 3 α = 29 - 2 \times \frac{11}{2} \Rightarrow 3 α = 18$

$\Rightarrow α = 6$ (maximum)

$∴$ Product of the smallest and the largest value of $α = \frac{11}{2} \times 6 = 33$ .

Q 11 :

Let ${}^{n}C_{r - 1} = 28$ , ${}^{n}C_{r} = 56$ and ${}^{n}C_{r + 1} = 70$ . Let A(4 cos t, 4 sin t), B(2 sin t, – 2 cos t) and C(3r – n, $r^{2}$ – n – 1) be the vertices of a triangle ABC, where t is a parameter. If ${(3 x - 1)}^{2} + {(3 y)}^{2} = α$ , is the locus of the centroid of triangle ABC, then $α$ equals [2025]

18
8
20
6

1

2

3

4

(3)

We have, ${}^{n}C_{r - 1} = 28$ , ${}^{n}C_{r} = 56$ and ${}^{n}C_{r + 1} = 70$

$\frac{{}^{n}C_{r - 1}}{{}^{n}C_{r}} = \frac{28}{56} \Rightarrow \frac{n!}{(r - 1)! (n - r + 1)!} \times \frac{r! (n - r)!}{n!} = \frac{1}{2}$

$\Rightarrow \frac{r}{(n - r + 1)} = \frac{1}{2} \Rightarrow 3 r = n + 1$ ... (i)

Also, $\frac{{}^{n}C_{r}}{{}^{n}C_{r + 1}} = \frac{56}{70} \Rightarrow \frac{r + 1}{n - r} = \frac{4}{5} \Rightarrow 9 r = 4 n - 5$ ... (ii)

From equations (i) and (ii), we get r = 3 and n = 8.

Now, A = (4 cos t, 4 sin t), B(2 sin t, – 2 cos t) and C(1, 0)

$∴$ Centroid of triangle is

$(\frac{4 c o s t + 2 \sin t + 1}{3}, \frac{4 \sin t - 2 \cos t}{3})$

Then, ${(3 x - 1)}^{2} + {(3 y)}^{2} = {(4 \cos t + 2 \sin t)}^{2} + {(4 \sin t - 2 \cos t)}^{2} {(3 x - 1)}^{2} + {(3 y)}^{2} = 20$

So, value of $α = 20$ .

Q 12 :

Let A(4, –2), B(1, 1) and C(9, –3) be the vertices of a triangle ABC. Then the maximum area of the parallelogram AFDE, formed with vertices D, E and F on the sides BC, CA and AB of the triangle ABC respectively, is __________.. [2025]

(3)

The maximum area of a parallelogram inscribed in a triangle is half the area of the triangle.

Now, Area of $△ A B C = \frac{1}{2}$ $| 4 (1 - (- 3)) + 1 (- 3 - (- 2)) + 9 (- 2 - 1) |$

$= \frac{1}{2}$ $| 16 - 1 - 27 |$ $= 6 sq . units$

$∴$ Area of parallelogram AFDE = 3 sq. units.

Q 13 :

Let A(6, 8), B(10 cos $α$ , – 10 sin $α$ ) and C(–10 sin $α$ , 10 cos $α$ ), be the vertices of a triangle. If L(a, 9) and G(h, k) be its orthocenter and centroid respectively, then (5a – 3h + 6k + 100 sin 2 $α$ ) is equal to __________. [2025]

(145)

We can observe that all the three points A, B, C lie on the circle $x^{2} + y^{2} = 100$ , so circumcentre is (0, 0). Since, centroid divides the line joining orthocentre and circumcentre in the ratio 2 : 1, then

$\frac{a + 0}{3} = h \Rightarrow a = 3 h$

and $\frac{9 + 0}{3} = k \Rightarrow k = 3$

Also, $\frac{6 + 10 \cos α - 10 \sin α}{3} = h$

$\Rightarrow 10 (\cos α - \sin α) = 3 h - 6$ ... (i)

and $\frac{8 + 10 \cos α - 10 \sin α}{3} = k$

$\Rightarrow 10 (\cos α - \sin α) = 3 k - 8 = 9 - 8 = 1$ ... (ii)

On squaring, 100 (1 – sin 2 $α$ ) = 1 $\Rightarrow$ 100 sin 2 $α$ = 99

From (i) and (ii), we get $h = \frac{7}{3}$

Now, 5a – 3h + 6k +100 sin 2 $α$

= 15h – 3h + 6k + 100 sin 2 $α$ = $12 \times \frac{7}{3} + 18 + 99$ = 145.

Q 14 :

Let $G (α, β)$ be the circumcentre of the triangle formed by the lines $4 x + 3 y = 69, 4 y - 3 x = 17$ and $x + 7 y = 61$ . Then ${(α - β)}^{2} + α + β$ is equal to [2023]

16
17
18
15

1

2

3

4

(2)

$Let A B C be a triangle with$

$A B : 4 x + 3 y = 69 \dots (i)$

$B C : 4 y - 3 x = 17 \dots (i i)$

$A C : x + 7 y = 61 \dots (i i i)$

Solving (i) and (ii), we get $B \equiv (9, 11)$

Solving (ii) and (iii), we get $C \equiv (5, 8)$

Solving (i) and (iii), we get $A \equiv (12, 7)$

Now, $A G^{2} = G B^{2}$

$\Rightarrow {(12 - α)}^{2} + {(7 - β)}^{2} = {(9 - α)}^{2} + {(11 - β)}^{2}$

$\Rightarrow 8 β - 6 α = 9 ...(iv)$

$Similarly, G C^{2} = G B^{2}$

$\Rightarrow {(5 - α)}^{2} + {(8 - β)}^{2} = {(9 - α)}^{2} + {(11 - β)}^{2}$

$\Rightarrow 8 α + 6 β = 113 \dots (v)$

Now solving (iv) and (v), we get

$α = \frac{17}{2}, β = \frac{15}{2}$

$∴$ ${(α - β)}^{2} + α + β = 17$

Q 15 :

Let R be the rectangle given by the lines $x = 0, x = 2, y = 0$ and $y = 5$ . Let $A (α, 0)$ and $B (0, β), α \in [0, 2]$ and $β \in [0, 5]$ , be such that the line segment AB divides the area of the rectangle R in the ratio 4 : 1. Then, the midpoint of AB lies on a [2023]

parabola
hyperbola
straight line
circle

1

2

3

4

(2)

$\frac{ar (A P Q R B)}{ar (∆ O A B)} = \frac{4}{1}$

Let M be the mid-point of AB.

$M (h, k) \equiv (\frac{α}{2}, \frac{β}{2})$

$\Rightarrow \frac{10 - \frac{1}{2} α β}{\frac{1}{2} α β} = 4 \Rightarrow \frac{5}{2} α β = 10$

$\Rightarrow α β = 4$

$\Rightarrow (2 h) (2 k) = 4$

$∴$ $Locus of M is x y = 1, which is a hyperbola.$

Q 16 :

If the point $(α, \frac{7 \sqrt{3}}{3})$ lies on the curve traced by the mid-points of the line segments of the lines $x \cos θ + y \sin θ = 7, θ \in (0, \frac{π}{2})$ between the coordinate axes, then $α$ is equal to [2023]

- 7
$- 7 \sqrt{3}$
$7 \sqrt{3}$
7

1

2

3

4

(4)

$Given line is x \cos θ + y \sin θ = 7$

$x -intercept = \frac{7}{\cos θ}, y -intercept = \frac{7}{\sin θ}$

Let locus of intercept be $(h, k)$ .

$(h, k) = (\frac{7}{2 \cos θ}, \frac{7}{2 \sin θ})$

$h = \frac{7}{2 \cos θ}, k = \frac{7}{2 \sin θ}$

$α = \frac{7}{2 \cos θ}, \frac{7 \sqrt{3}}{3} = \frac{7}{2 \sin θ}$

$\cos θ = \frac{7}{2 α}, \sin θ = \frac{\sqrt{3}}{2}$

$\cos^{2} θ + \sin^{2} θ = 1$

$\Rightarrow \frac{49}{4 α^{2}} + \frac{3}{4} = 1 \Rightarrow \frac{49}{4 α^{2}} = \frac{1}{4}$

$α^{2} = 49$

$∴$ $α = 7 (As θ \in (0, \frac{π}{2}))$

Q 17 :

Let A(1, 2) and C(−3, −6) be two diagonally opposite vertices of a rhombus, whose sides AD and BC are parallel to the line $7 x - y = 14$ If B $(α, β)$ and $D (γ, δ)$ are the other two vertices, then $| α + β + γ + δ |$ is equal to. [2026]

6
1
3
9

1

2

3

4

(1)

$Given the points of B and D are (α, β) and (γ, δ)$ $and midpoint of A and C is (- 1, - 2)$

$So \frac{α + γ}{2} = - 1 and \frac{β + δ}{2} = - 2$

$| α + γ + β + δ |$ $= 6$

Q 18 :

Among the statements

(S1): If A(5,−1) and B(−2,3) are two vertices of a triangle, whose orthocentre is (0,0), then its third vertex is (−4,−7).

(S2): If positive numbers 2a,b,c are three consecutive terms of an A.P., then the lines ax+by+c=0 are concurrent at (2,−2). [2026]

both are correct
both are incorrect
only (S2) is correct
only (S1) is correct

1

2

3

4

(1)

$Solution of statement-1$

$m_{A O} \cdot m_{B C} = - 1$

$B (- 2, 3)$

$\Rightarrow 5 h - k + 13 = 0 . . . (1)$

$& m_{A B} \cdot m_{O C} = - 1$

$\Rightarrow 4 k = 7 h . . . (2)$

$\Rightarrow third vertex is (- 4, - 7)$

$∴$ $Statement 1 is correct.$

$Solution of statement-2$

$2 a, b, c \to A.P.$

$b = \frac{2 a + c}{2}$

$\Rightarrow 2 a - 2 b + c = 0$

$∵$ $lines a x + b y + c = 0 are concurrent then$

$\frac{x}{2} = \frac{y}{- 2} = \frac{1}{1}$

$x = 2 and y = - 2$

$∴$ $Point of concurrency is (2, - 2)$

$∴$ $Statement 2 is correct.$

Q 19 :

Let ABC be an equilateral triangle with orthocenter at the origin and the side BC on the line $x + 2 \sqrt{2} y = 4$ If the co-ordinates of the vertex A are $(α, β)$ then the greatest integer less than or equal to $|| α + \sqrt{2} β ||$ is. [2026]

2
5
4
3

1

2

3

4

(3)

$∵$ $m_{B C} \cdot m_{A D} = - 1$

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

$\Rightarrow β = 2 \sqrt{2} α . . . (1)$

$∵$ $O D =$ $| \frac{- 4}{\sqrt{1 + 8}} |$ $= \frac{4}{3} \Rightarrow A O = \frac{8}{3}$

$So A D = \frac{8}{3} + \frac{4}{3} = 4$

$\Rightarrow \frac{| α + 2 \sqrt{2} β - 4 |}{3} = 4 \Rightarrow α = \frac{16}{9} or - \frac{8}{9}$

${∵ A (α, β) & (0, 0) lie on same side of given line}$

$∴$ $(α, β) = (\frac{16}{9}, \frac{32 \sqrt{2}}{9}) (Rejected)$

$so (α, β) = (- \frac{8}{9}, - \frac{16 \sqrt{2}}{9})$

$= [| α + \sqrt{2} β |] = [| \frac{- 8 - 32}{9} |] = 4$

Topic Question Set

Q 1 :

Let a variable line of slope m > 0 passing through the point (4, –9) intersect the coordinate axes at the points A and B. The minimum value of the sum of the distances of A and B from the origin is [2024]

Q 2 :

If the locus of the point, whose distance from the point (2, 1) and (1, 3) are in the ratio 5 : 4, is $a x^{2} + b y^{2} + c x y + d x + e y + 170 = 0$ , then the value of $a^{2} + 2 b + 3 c + 4 d + e$ is equal to : [2024]

Q 3 :

If the line segment joining the points (5, 2) and (2, a) subtends an angle $\frac{π}{4}$ at the origin, then the absolute value of the product of all possible values of a is: [2024]

Q 4 :

A ray of light coming from the point P(1, 2) gets reflected from the point Q on the x-axis and then passes through the point R(4, 3). If the point S(h, k) is such that PQRS is a parallelogram, then $h k^{2}$ is equal to: [2024]

Q 5 :

Two vertices of a triangle ABC are A(3, –1) and B(–2, 3) and its orthocentre is P(1, 1). If the coordinates of the point C are $(α, β)$ and the centre of the circle circumscribing the triangle PAB is (h, k), then the value of $(α + β) + 2 (h + k)$ equals [2024]

Q 6 :

Let $α, β, γ, δ \in Z$ and let $A (α, β), B (1, 0), C (γ, δ)$ and D(1, 2) be the vertices of a parallelogram ABCD. If $A B = \sqrt{10}$ and the points A and C lie on the line 3y = 2x + 1, then $2 (α + β + γ + δ)$ is equal to [2024]

Q 7 :

Let $A (- 2, - 1), B (1, 0), C (α, β)$ and $D (γ, δ)$ be the vertices of a parallelogram ABCD. If the point C lies on 2x – y = 5 and the point D lies on 3x – 2y = 6, then the value of $| α + β + γ + δ |$ is equal to __________. [2024]

Q 8 :

Let the three sides of a triangle are on the lines 4x – 7y + 10 = 0, x + y = 5 and 7x + 4y = 15. Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines x = 0, y = 0 and x + y = 1 is [2025]

Q 9 :

Let the area of a $△$ PQR with vertices P(5, 4), Q(–2, 4) and R(a, b) be 35 square units. If its orthocenter and centroid are $O (2, \frac{14}{5})$ and C(c, d) respectively, then c + 2d is equal to [2025]

Q 10 :

Let the points $(\frac{11}{2}, α)$ lie on or inside the triangle with sides x + y = 11, x + 2y = 16 and 2x + 3y = 29. Then the product of the smallest and the largest values of $α$ is equal to : [2025]

Q 12 :

Let A(4, –2), B(1, 1) and C(9, –3) be the vertices of a triangle ABC. Then the maximum area of the parallelogram AFDE, formed with vertices D, E and F on the sides BC, CA and AB of the triangle ABC respectively, is __________.. [2025]

Q 13 :

Let A(6, 8), B(10 cos $α$ , – 10 sin $α$ ) and C(–10 sin $α$ , 10 cos $α$ ), be the vertices of a triangle. If L(a, 9) and G(h, k) be its orthocenter and centroid respectively, then (5a – 3h + 6k + 100 sin 2 $α$ ) is equal to __________. [2025]

Q 14 :

Let $G (α, β)$ be the circumcentre of the triangle formed by the lines $4 x + 3 y = 69, 4 y - 3 x = 17$ and $x + 7 y = 61$ . Then ${(α - β)}^{2} + α + β$ is equal to [2023]

Q 15 :

Let R be the rectangle given by the lines $x = 0, x = 2, y = 0$ and $y = 5$ . Let $A (α, 0)$ and $B (0, β), α \in [0, 2]$ and $β \in [0, 5]$ , be such that the line segment AB divides the area of the rectangle R in the ratio 4 : 1. Then, the midpoint of AB lies on a [2023]

Q 16 :

If the point $(α, \frac{7 \sqrt{3}}{3})$ lies on the curve traced by the mid-points of the line segments of the lines $x \cos θ + y \sin θ = 7, θ \in (0, \frac{π}{2})$ between the coordinate axes, then $α$ is equal to [2023]

Q 17 :

Let A(1, 2) and C(−3, −6) be two diagonally opposite vertices of a rhombus, whose sides AD and BC are parallel to the line $7 x - y = 14$ If B $(α, β)$ and $D (γ, δ)$ are the other two vertices, then $| α + β + γ + δ |$ is equal to. [2026]

Q 18 :

Among the statements

(S1): If A(5,−1) and B(−2,3) are two vertices of a triangle, whose orthocentre is (0,0), then its third vertex is (−4,−7).

(S2): If positive numbers 2a,b,c are three consecutive terms of an A.P., then the lines ax+by+c=0 are concurrent at (2,−2). [2026]

Q 19 :

Let ABC be an equilateral triangle with orthocenter at the origin and the side BC on the line $x + 2 \sqrt{2} y = 4$ If the co-ordinates of the vertex A are $(α, β)$ then the greatest integer less than or equal to $|| α + \sqrt{2} β ||$ is. [2026]

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Topic Question Set

Q 1 : Let a variable line of slope m > 0 passing through the point (4, –9) intersect the coordinate axes at the points A and B. The minimum value of the sum of the distances of A and B from the origin is [2024]

Q 2 : If the locus of the point, whose distance from the point (2, 1) and (1, 3) are in the ratio 5 : 4, is ax2+by2+cxy+dx+ey+170=0, then the value of a2+2b+3c+4d+e is equal to : [2024]

Q 3 : If the line segment joining the points (5, 2) and (2, a) subtends an angle π4 at the origin, then the absolute value of the product of all possible values of a is: [2024]

Q 4 : A ray of light coming from the point P(1, 2) gets reflected from the point Q on the x-axis and then passes through the point R(4, 3). If the point S(h, k) is such that PQRS is a parallelogram, then hk2 is equal to: [2024]

Q 5 : Two vertices of a triangle ABC are A(3, –1) and B(–2, 3) and its orthocentre is P(1, 1). If the coordinates of the point C are (α,β) and the centre of the circle circumscribing the triangle PAB is (h, k), then the value of (α+β)+2(h+k) equals [2024]

Q 6 : Let α,β,γ, δ∈Z and let A(α,β),B(1,0),C(γ,δ) and D(1, 2) be the vertices of a parallelogram ABCD. If AB=10 and the points A and C lie on the line 3y = 2x + 1, then 2(α+β+γ+δ) is equal to [2024]

Q 7 : Let A(–2,–1),B(1,0),C(α,β) and D(γ,δ) be the vertices of a parallelogram ABCD. If the point C lies on 2x – y = 5 and the point D lies on 3x – 2y = 6, then the value of |α+β+γ+δ| is equal to __________. [2024]

Q 8 : Let the three sides of a triangle are on the lines 4x – 7y + 10 = 0, x + y = 5 and 7x + 4y = 15. Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines x = 0, y = 0 and x + y = 1 is [2025]

Q 9 : Let the area of a △PQR with vertices P(5, 4), Q(–2, 4) and R(a, b) be 35 square units. If its orthocenter and centroid are O(2,145) and C(c, d) respectively, then c + 2d is equal to [2025]

Q 10 : Let the points (112,α) lie on or inside the triangle with sides x + y = 11, x + 2y = 16 and 2x + 3y = 29. Then the product of the smallest and the largest values of α is equal to : [2025]

Q 11 : Let Cr–1n=28, Crn=56 and Cr+1n=70. Let A(4 cos t, 4 sin t), B(2 sin t, – 2 cos t) and C(3r – n, r2– n – 1) be the vertices of a triangle ABC, where t is a parameter. If (3x–1)2+(3y)2=α, is the locus of the centroid of triangle ABC, then α equals [2025]

Q 12 : Let A(4, –2), B(1, 1) and C(9, –3) be the vertices of a triangle ABC. Then the maximum area of the parallelogram AFDE, formed with vertices D, E and F on the sides BC, CA and AB of the triangle ABC respectively, is __________.. [2025]

Q 13 : Let A(6, 8), B(10 cos α, – 10 sin α) and C(–10 sin α, 10 cos α), be the vertices of a triangle. If L(a, 9) and G(h, k) be its orthocenter and centroid respectively, then (5a – 3h + 6k + 100 sin 2α) is equal to __________. [2025]

Q 14 : Let G(α,β) be the circumcentre of the triangle formed by the lines 4x+3y=69,4y-3x=17 and x+7y=61. Then (α-β)2+α+β is equal to [2023]

Q 15 : Let R be the rectangle given by the lines x=0,x=2,y=0 and y=5. Let A(α,0) and B(0,β),α∈[0,2] and β∈[0,5], be such that the line segment AB divides the area of the rectangle R in the ratio 4 : 1. Then, the midpoint of AB lies on a [2023]

Q 16 : If the point (α,733) lies on the curve traced by the mid-points of the line segments of the lines xcosθ+ysinθ=7,θ∈(0,π2) between the coordinate axes, then α is equal to [2023]

Q 17 : Let A(1, 2) and C(−3, −6) be two diagonally opposite vertices of a rhombus, whose sides AD and BC are parallel to the line 7x−y=14 If B (α,β) and D(γ,δ) are the other two vertices, then |α+β+γ+δ| is equal to. [2026]

Q 18 : Among the statements (S1): If A(5,−1) and B(−2,3) are two vertices of a triangle, whose orthocentre is (0,0), then its third vertex is (−4,−7). (S2): If positive numbers 2a,b,c are three consecutive terms of an A.P., then the lines ax+by+c=0 are concurrent at (2,−2). [2026]

Q 19 : Let ABC be an equilateral triangle with orthocenter at the origin and the side BC on the line x+22y=4 If the co-ordinates of the vertex A are (α,β) then the greatest integer less than or equal to |α+2β| is. [2026]

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Q 1 :

Let a variable line of slope m > 0 passing through the point (4, –9) intersect the coordinate axes at the points A and B. The minimum value of the sum of the distances of A and B from the origin is [2024]

Q 2 :

If the locus of the point, whose distance from the point (2, 1) and (1, 3) are in the ratio 5 : 4, is $a x^{2} + b y^{2} + c x y + d x + e y + 170 = 0$ , then the value of $a^{2} + 2 b + 3 c + 4 d + e$ is equal to : [2024]

Q 3 :

If the line segment joining the points (5, 2) and (2, a) subtends an angle $\frac{π}{4}$ at the origin, then the absolute value of the product of all possible values of a is: [2024]

Q 4 :

A ray of light coming from the point P(1, 2) gets reflected from the point Q on the x-axis and then passes through the point R(4, 3). If the point S(h, k) is such that PQRS is a parallelogram, then $h k^{2}$ is equal to: [2024]

Q 5 :

Two vertices of a triangle ABC are A(3, –1) and B(–2, 3) and its orthocentre is P(1, 1). If the coordinates of the point C are $(α, β)$ and the centre of the circle circumscribing the triangle PAB is (h, k), then the value of $(α + β) + 2 (h + k)$ equals [2024]

Q 6 :

Let $α, β, γ, δ \in Z$ and let $A (α, β), B (1, 0), C (γ, δ)$ and D(1, 2) be the vertices of a parallelogram ABCD. If $A B = \sqrt{10}$ and the points A and C lie on the line 3y = 2x + 1, then $2 (α + β + γ + δ)$ is equal to [2024]

Q 7 :

Let $A (- 2, - 1), B (1, 0), C (α, β)$ and $D (γ, δ)$ be the vertices of a parallelogram ABCD. If the point C lies on 2x – y = 5 and the point D lies on 3x – 2y = 6, then the value of $| α + β + γ + δ |$ is equal to __________. [2024]

Q 8 :

Let the three sides of a triangle are on the lines 4x – 7y + 10 = 0, x + y = 5 and 7x + 4y = 15. Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines x = 0, y = 0 and x + y = 1 is [2025]

Q 9 :

Let the area of a $△$ PQR with vertices P(5, 4), Q(–2, 4) and R(a, b) be 35 square units. If its orthocenter and centroid are $O (2, \frac{14}{5})$ and C(c, d) respectively, then c + 2d is equal to [2025]

Q 10 :

Let the points $(\frac{11}{2}, α)$ lie on or inside the triangle with sides x + y = 11, x + 2y = 16 and 2x + 3y = 29. Then the product of the smallest and the largest values of $α$ is equal to : [2025]

Q 12 :

Let A(4, –2), B(1, 1) and C(9, –3) be the vertices of a triangle ABC. Then the maximum area of the parallelogram AFDE, formed with vertices D, E and F on the sides BC, CA and AB of the triangle ABC respectively, is __________.. [2025]

Q 13 :

Let A(6, 8), B(10 cos $α$ , – 10 sin $α$ ) and C(–10 sin $α$ , 10 cos $α$ ), be the vertices of a triangle. If L(a, 9) and G(h, k) be its orthocenter and centroid respectively, then (5a – 3h + 6k + 100 sin 2 $α$ ) is equal to __________. [2025]

Q 14 :

Let $G (α, β)$ be the circumcentre of the triangle formed by the lines $4 x + 3 y = 69, 4 y - 3 x = 17$ and $x + 7 y = 61$ . Then ${(α - β)}^{2} + α + β$ is equal to [2023]

Q 15 :

Let R be the rectangle given by the lines $x = 0, x = 2, y = 0$ and $y = 5$ . Let $A (α, 0)$ and $B (0, β), α \in [0, 2]$ and $β \in [0, 5]$ , be such that the line segment AB divides the area of the rectangle R in the ratio 4 : 1. Then, the midpoint of AB lies on a [2023]

Q 16 :

If the point $(α, \frac{7 \sqrt{3}}{3})$ lies on the curve traced by the mid-points of the line segments of the lines $x \cos θ + y \sin θ = 7, θ \in (0, \frac{π}{2})$ between the coordinate axes, then $α$ is equal to [2023]

Q 17 :

Let A(1, 2) and C(−3, −6) be two diagonally opposite vertices of a rhombus, whose sides AD and BC are parallel to the line $7 x - y = 14$ If B $(α, β)$ and $D (γ, δ)$ are the other two vertices, then $| α + β + γ + δ |$ is equal to. [2026]

Q 18 :

Among the statements

(S1): If A(5,−1) and B(−2,3) are two vertices of a triangle, whose orthocentre is (0,0), then its third vertex is (−4,−7).

(S2): If positive numbers 2a,b,c are three consecutive terms of an A.P., then the lines ax+by+c=0 are concurrent at (2,−2). [2026]

Q 19 :

Let ABC be an equilateral triangle with orthocenter at the origin and the side BC on the line $x + 2 \sqrt{2} y = 4$ If the co-ordinates of the vertex A are $(α, β)$ then the greatest integer less than or equal to $|| α + \sqrt{2} β ||$ is. [2026]