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

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

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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]

0%

(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]

0%

(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

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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

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(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

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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 $(h, k)$ .

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

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

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

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

$\cos^{2} θ + \sin^{2} θ = 1$

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

$α^{2} = 49$

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

Topic Question Set

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]

0%

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]

0%

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]

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