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

Q 11 :
If $x^{2} - y^{2} + 2 h x y + 2 g x + 2 f y + c = 0$ is the locus of a point, which moves such that it is always equidistant from the lines x + 2y + 7 = 0 and 2x – y + 8 = 0, then the value of g + c + h – f equals [2024]

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14

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(4)

Let point P(x, y) be equidistant from the given lines.

$∴ \frac{x + 2 y + 7}{\sqrt{5}} = \pm \frac{2 x - y + 8}{\sqrt{5}}$

$\Rightarrow {(x + 2 y + 7)}^{2} = {(2 x - y + 8)}^{2}$

$\Rightarrow x^{2} + 4 y^{2} + 49 + 4 x y + 28 y + 14 x = 4 x^{2} + y^{2} + 64 - 4 x y - 16 y + 32 x$

$\Rightarrow 3 x^{2} - 3 y^{2} - 8 x y + 18 x - 44 y + 15 = 0$

$\Rightarrow x^{2} - y^{2} - \frac{8}{3} x y + 6 x - \frac{44}{3} y + 5 = 0$ ... (i)

This is the locus of the point P(x, y).

Now, compare equation (i) with given equation of locus, we get

$2 h = \frac{- 8}{3} \Rightarrow h = \frac{- 4}{3}$ ,

$2 g = 6 \Rightarrow g = 3$ ,

$2 f = \frac{- 44}{3} \Rightarrow \frac{- 22}{3}$ ,

and c = 5

$∴ g + c + h - f = 3 + 5 - \frac{4}{3} + \frac{22}{3} = 14$ .

Q 12 :
Let A (a, b), B(3, 4) and C(–6, –8) respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point P(2a + 3, 7b + 5) from the line 2x + 3y – 4 = 0 measured parallel to the line x – 2y – 1 = 0 is [2024]

$\frac{17 \sqrt{5}}{6}$
$\frac{\sqrt{5}}{17}$
$\frac{17 \sqrt{5}}{7}$
$\frac{15 \sqrt{5}}{7}$

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(3)

Given the A (a, b), B(3, 4) and C(–6, –8) respectively the centroid, circumcentre and orthocentre of a triangle.

We know that centroid divides circumcentre and orthocentre internally in the ratio 1 : 2

$∴ a = \frac{- 6 + 6}{2}, b = \frac{- 8 + 8}{3}$

$\Rightarrow$ a = 0, b = 0

Therefore, the coordinates of P are (3, 5).

Also, $l$ : 2x + 3y – 4 = 0 (Given) ... (i)

Equation of the line passing through P(3, 5) and parallel to the line x – 2y – 1 = 0 is

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

$\Rightarrow 2 y - 10 = x - 3$

$\Rightarrow x - 2 y + 7 = 0$ ... (ii)

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

$x = \frac{- 13}{7}, y = \frac{18}{7}$

$∴$ Distance between (3, 5) and $(\frac{- 13}{7}, \frac{18}{7})$ is

$\sqrt{{(3 + \frac{13}{7})}^{2} + {(5 - \frac{18}{7})}^{2}} = \sqrt{\frac{1156}{49} + \frac{289}{49}} \sqrt{\frac{1445}{49}} = \frac{17 \sqrt{5}}{7}$

Q 13 :
Consider a triangle ABC having the vertices A(1, 2), $B (α, β)$ and $C (γ, δ)$ and angles $∠ A B C = \frac{π}{6}$ and $∠ B A C = \frac{2 π}{3}$ . If the points B and C lie on the line y = x + 4, then $α^{2} + γ^{2}$ is equal to __________ [2024]

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(14)

$P = \frac{| - 1 + 2 - 4 |}{\sqrt{1^{2} + 1^{2}}} = \frac{3}{\sqrt{2}}$

$\sin (\frac{π}{6}) = \frac{3 / \sqrt{2}}{A B} \Rightarrow \frac{3}{\sqrt{2} A B} = \frac{1}{2} \Rightarrow A B = \frac{6}{\sqrt{2}}$

$\Rightarrow {(α - 1)}^{2} + {(α + 4 - 2)}^{2} = {(\frac{6}{\sqrt{2}})}^{2}$

$\Rightarrow α^{2} + 1 - 2 α + α^{2} + 4 + 4 α = 18$

$\Rightarrow 2 α^{2} + 2 α - 13 = 0 \to α and γ$ satisfy same equation

$∴$ $α$ and $γ$ are the roots of the equation $2 x^{2} + 2 x - 13 = 0$

Sum of roots = $α + γ = \frac{- 2}{2} = - 1$

Product of roots = $α γ = \frac{- 13}{2}$

Now, $α^{2} + γ^{2} = {(α + γ)}^{2} - 2 α γ$

$= {(- 1)}^{2} - 2 (\frac{- 13}{2}) = 1 + 13 = 14$ .

Q 14 :
Let a ray of light passing through the point (3, 10) reflects on the line 2x + y = 6 and the reflected ray passes through the point (7, 2). If the equation of the incident ray is ax + by + 1 = 0, then $a^{2} + b^{2} + 3 a b$ is equal to __________ . [2024]

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(1)

$\frac{x - 3}{2} = \frac{y - 10}{1} = \frac{- 2 (2 \times 3 + 10 - 6)}{4 + 1} = - 4$

$\Rightarrow x = - 5, y = 6$

$∴$ A'(–5, 6) and B(7, 2)

$∴$ Equation of line A'B is

$\Rightarrow y - 6 = \frac{2 - 6}{7 + 5} (x + 5)$

$\Rightarrow y - 6 = - \frac{1}{3} (x + 5)$

$\Rightarrow 3 y - 18 = - x - 5 \Rightarrow x + 3 y = 13$

and 2x + y = 6 (Given line).

On solving, we get y = 4, x = 1

$∴ Q \equiv (1, 4)$

Equation of line AQ is

$y - 10 = \frac{10 - 4}{3 - 1} (x - 3) \Rightarrow y - 10 = 3 (x - 3)$

$\Rightarrow y - 10 = 3 x - 9 \Rightarrow 3 x - y + 1 = 0$

On comparing with given equation ax + by + 1 = 0, we get a = 3, b = –1

Hence, $a^{2} + b^{2} + 3 a b = 9 + 1 + 3 (3) (- 1) = 9 + 1 - 9 = 1$ .

Q 15 :
Let ABC be an isosceles triangle in which A is at (–1, 0), $∠ A = \frac{2 π}{3}$ , AB = AC and B is on the positive x-axis. If $B C = 4 \sqrt{3}$ and the line BC intersects the line y = x + 3 at $(α, β)$ , then $\frac{β^{4}}{α^{2}}$ is __________. [2024]

0%

(36)

We have, $\frac{\sin 30 °}{A C} = \frac{\sin 120 °}{B C}$

$\Rightarrow A C = 4 \sqrt{3} \times \frac{1}{2} \times \frac{2}{\sqrt{3}}$

$\Rightarrow A C = 4 = A B$

$\Rightarrow A B^{2} = {(x + 1)}^{2} \Rightarrow x + 1 = \pm 4$

But B is on positive x-axis

$∴$ Coordinates of B = (3, 0)

Now, $A C^{2} = 16$

$\Rightarrow {(p + 1)}^{2} + q^{2} = 16$ ... (i)

Also, $B C^{2} = 48$

$\Rightarrow {(p - 3)}^{2} + q^{2} = 48$ ... (ii)

Solving (i) and (ii), we get

$p = - 3, q = 2 \sqrt{3}$

So, coordinates of $C = (- 3, 2 \sqrt{3})$

$∴$ Equation of line BC is, $y - 0 = \frac{2 \sqrt{3} - 0}{- 3 - 3} (x - 3)$

$\Rightarrow x + \sqrt{3} y = 3$ ... (iii)

Now, point of intersection of line (iii) and x + 3 = y is

$(3 (\sqrt{3} - 2), 3 (\sqrt{3} - 1))$

$\Rightarrow α = 3 (\sqrt{3} - 2), β = 3 (\sqrt{3} - 1)$

$∴$ $\frac{β^{4}}{α^{2}} = \frac{3^{4} {(3 + 1 - 2 \sqrt{3})}^{2}}{3^{2} {(\sqrt{3} - 2)}^{2}} = \frac{9 (2 {(2 - \sqrt{3}))}^{2}}{{(2 - \sqrt{3})}^{2}} = 9 \times 4 = 36$ .

Q 16 :
If the sum of squares of all real values of $α$ , for which the lines 2x – y + 3 = 0, 6x + 3y + 1 = 0 and $α x + 2 y - 2 = 0$ do not form a triangle is p, then the greatest integer less than or equal to p is __________. [2024]

0%

(32)

We have, 2x – y + 3 = 0 ... (i)

6x + 3y + 1 = 0 ... (ii)

$α x + 2 y - 2 = 0$ ... (iii)

Case I. If the lines are concurrent then they do not form a triangle

$∴$ $| \begin{matrix} 2 & - 1 & 3 \\ 6 & 3 & 1 \\ α & 2 & - 2 \end{matrix} |$ $= 0$

$\Rightarrow α (- 1 - 9) - 2 (2 - 18) - 2 (6 + 6) = 0$

$\Rightarrow - 10 α + 32 - 24 = 0 \Rightarrow 10 α = 8 \Rightarrow α = \frac{4}{5}$

Case II. If the lines are parallel then they do not form a triangle.

If the lines (i) and (iii) are parallel,

$∴ \frac{2}{α} = \frac{- 1}{2} \neq \frac{- 3}{2} \Rightarrow α = - 4$

If the lines (ii) and (iii) are parallel,

$∴ \frac{6}{α} = \frac{3}{2} \neq \frac{- 1}{2} \Rightarrow α = 4$

$∴$ Sum of squares of all real values of $α$

$= {(\frac{4}{5})}^{2} + {(- 4)}^{2} + {(4)}^{2} = \frac{16}{25} + 16 + 16 = \frac{16}{25} + 32$

$[P] = [32 + \frac{16}{25}] = 32$ .

Q 17 :
Let for any three distinct consecutive terms a, b, c of an A.P., the lines ax + by + c = 0 be concurrent at the point P and $Q (α, β)$ be a point such that the system of equations x + y + z = 6, 2x + 5y + $α$ z = $β$ and x + 2y + 3z = 4, has infinitely many solutions. Then ${(P Q)}^{2}$ is equal to __________. [2024]

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(113)

a, b, c are in A.P. $\Rightarrow$ 2b = a + c $\Rightarrow$ a – 2b + c = 0 and ax + by + c = 0 are concurrent.

Two lines $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ are concurrent if

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} ∴ \frac{1}{x} = \frac{- 2}{y} = \frac{1}{1} = 0$

$\Rightarrow x = 1 and y = - 2$

So, P(1, –2) is the point of concurrency.

Now, x + y + z = 6 ... (i)

2x + 5y + $α$ z = $β$ ... (ii)

x + 2y + 3z = 4 ... (iii)

has infinitely many solution.

On solving these equation, we have

x + y = 6 – z, x + 2y = 4 – 3z

On solving these two, we get –y = 2 + 2z

$\Rightarrow$ y = –2(1 + z) $\Rightarrow$ x = 6 – z + 2 + 2z $\Rightarrow$ x = 8 + z

From (ii), we get 2(8 + z) + 5(–2(1 + z)) + $α$ z = $β$

$\Rightarrow 6 - 8 z + α z = β$

$\Rightarrow z (α - 8) = β - 6$

For infinitely many solution, $α = 8, β = 6$

$| P Q |^{2}$ $= {(\sqrt{{(8 - 1)}^{2} + {(6 + 2)}^{2}})}^{2} = 113$

Topic Question Set

Q 11 :
If $x^{2} - y^{2} + 2 h x y + 2 g x + 2 f y + c = 0$ is the locus of a point, which moves such that it is always equidistant from the lines x + 2y + 7 = 0 and 2x – y + 8 = 0, then the value of g + c + h – f equals [2024]

Q 12 :
Let A (a, b), B(3, 4) and C(–6, –8) respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point P(2a + 3, 7b + 5) from the line 2x + 3y – 4 = 0 measured parallel to the line x – 2y – 1 = 0 is [2024]

Q 13 :
Consider a triangle ABC having the vertices A(1, 2), $B (α, β)$ and $C (γ, δ)$ and angles $∠ A B C = \frac{π}{6}$ and $∠ B A C = \frac{2 π}{3}$ . If the points B and C lie on the line y = x + 4, then $α^{2} + γ^{2}$ is equal to __________ [2024]

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Q 14 :
Let a ray of light passing through the point (3, 10) reflects on the line 2x + y = 6 and the reflected ray passes through the point (7, 2). If the equation of the incident ray is ax + by + 1 = 0, then $a^{2} + b^{2} + 3 a b$ is equal to __________ . [2024]

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Q 15 :
Let ABC be an isosceles triangle in which A is at (–1, 0), $∠ A = \frac{2 π}{3}$ , AB = AC and B is on the positive x-axis. If $B C = 4 \sqrt{3}$ and the line BC intersects the line y = x + 3 at $(α, β)$ , then $\frac{β^{4}}{α^{2}}$ is __________. [2024]

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Q 16 :
If the sum of squares of all real values of $α$ , for which the lines 2x – y + 3 = 0, 6x + 3y + 1 = 0 and $α x + 2 y - 2 = 0$ do not form a triangle is p, then the greatest integer less than or equal to p is __________. [2024]

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Topic Question Set

Q 11 : If x2 – y2 + 2hxy + 2gx + 2fy + c = 0 is the locus of a point, which moves such that it is always equidistant from the lines x + 2y + 7 = 0 and 2x – y + 8 = 0, then the value of g + c + h – f equals [2024]

Q 12 : Let A (a, b), B(3, 4) and C(–6, –8) respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point P(2a + 3, 7b + 5) from the line 2x + 3y – 4 = 0 measured parallel to the line x – 2y – 1 = 0 is [2024]

Q 13 : Consider a triangle ABC having the vertices A(1, 2), B(α, β) and C(γ, δ) and angles ∠ABC = π6 and ∠BAC = 2π3. If the points B and C lie on the line y = x + 4, then α2 + γ2 is equal to __________ [2024]

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Q 14 : Let a ray of light passing through the point (3, 10) reflects on the line 2x + y = 6 and the reflected ray passes through the point (7, 2). If the equation of the incident ray is ax + by + 1 = 0, then a2 + b2 + 3ab is equal to __________ . [2024]

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Q 15 : Let ABC be an isosceles triangle in which A is at (–1, 0), ∠A = 2π3, AB = AC and B is on the positive x-axis. If BC = 43 and the line BC intersects the line y = x + 3 at (α, β), then β4α2 is __________. [2024]

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Q 16 : If the sum of squares of all real values of α, for which the lines 2x – y + 3 = 0, 6x + 3y + 1 = 0 and αx + 2y – 2 = 0 do not form a triangle is p, then the greatest integer less than or equal to p is __________. [2024]

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Q 11 :
If $x^{2} - y^{2} + 2 h x y + 2 g x + 2 f y + c = 0$ is the locus of a point, which moves such that it is always equidistant from the lines x + 2y + 7 = 0 and 2x – y + 8 = 0, then the value of g + c + h – f equals [2024]

Q 12 :
Let A (a, b), B(3, 4) and C(–6, –8) respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point P(2a + 3, 7b + 5) from the line 2x + 3y – 4 = 0 measured parallel to the line x – 2y – 1 = 0 is [2024]

Q 13 :
Consider a triangle ABC having the vertices A(1, 2), $B (α, β)$ and $C (γ, δ)$ and angles $∠ A B C = \frac{π}{6}$ and $∠ B A C = \frac{2 π}{3}$ . If the points B and C lie on the line y = x + 4, then $α^{2} + γ^{2}$ is equal to __________ [2024]

Q 14 :
Let a ray of light passing through the point (3, 10) reflects on the line 2x + y = 6 and the reflected ray passes through the point (7, 2). If the equation of the incident ray is ax + by + 1 = 0, then $a^{2} + b^{2} + 3 a b$ is equal to __________ . [2024]

Q 15 :
Let ABC be an isosceles triangle in which A is at (–1, 0), $∠ A = \frac{2 π}{3}$ , AB = AC and B is on the positive x-axis. If $B C = 4 \sqrt{3}$ and the line BC intersects the line y = x + 3 at $(α, β)$ , then $\frac{β^{4}}{α^{2}}$ is __________. [2024]

Q 16 :
If the sum of squares of all real values of $α$ , for which the lines 2x – y + 3 = 0, 6x + 3y + 1 = 0 and $α x + 2 y - 2 = 0$ do not form a triangle is p, then the greatest integer less than or equal to p is __________. [2024]