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 X(3r – n, $t^{2}$ – 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 sin t) and C(1, 0)

$∴$ Centroid of triangle is

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

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 in 2 $α$ ) is equal to __________. [2025]

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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). Sice, 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 $\frac{8 + 10 \cos α - 10 \sin α}{3} = k$ ... (ii)

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

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.

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]

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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 in 2 $α$ ) is equal to __________. [2025]

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Topic Question Set

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 X(3r – n, t2– 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]

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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 in 2α) is equal to __________. [2025]

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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 in 2 $α$ ) is equal to __________. [2025]