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]

0%

(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 25 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 , 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 $α = \frac{11}{2} \times 6 = 33$ .

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]

0%

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 25 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]

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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]

0%

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 25 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]

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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 25 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]