JEE MAIN Rectangular Cartesian Coordinates Previous Year Questions (PYQs) – Page 2 of 3
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The locus of the point of intersection of the lines $ \sqrt{2}x - y + 4\sqrt{2}k = 0$ and $\sqrt{2}kx + ky - 4\sqrt{2} = 0$ $(k$ is any non-zero real parameter$)$, is :
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Solution
A square, of each side $2$, lies above the $x$-axis and has one vertex at the origin.
If one of the sides passing through the origin makes an angle $30^\circ$ with the positive direction of the $x$-axis,
then the sum of the $x$-coordinates of the vertices of the square is:
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Solution
Let $P_n = \alpha^n + \beta^n$, $n \in \mathbb{N}$.
If $P_{10} = 123$, $P_9 = 76$, $P_8 = 47$ and $P_1 = 1$, then the quadratic equation having roots $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$ is:
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If a $\Delta $ABC has vertices A(–1, 7), B(–7, 1) and C(5, –5), then its orthocentre has coordinates :
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The distance between the two points A and A' which lie on y = 2 such that both the line segments AB and A' B (where B is the point (2, 3)) subtend angle ${\pi \over 4}$ at the origin, is equal to :
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Let a1, a2, a3, ..... be an A.P. If ${{{a_1} + {a_2} + .... + {a_{10}}} \over {{a_1} + {a_2} + .... + {a_p}}} = {{100} \over {{p^2}}}$, p $\ne$ 10, then ${{{a_{11}}} \over {{a_{10}}}}$ is equal to :
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The distance of line $3y - 2z - 1 = 0 = 3x - z + 4$ from the point (2, $-$1, 6) is :
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Let the triangle PQR be the image of the triangle with vertices $(1,3),(3,1)$ and $(2,4)$ in the line $x+2 y=2$. If the centroid of $\triangle \mathrm{PQR}$ is the point $(\alpha, \beta)$, then $15(\alpha-\beta)$ is equal to :
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The distance of the point $(7,10,11)$ from the line $\dfrac{x-4}{1}=\dfrac{y-4}{0}=\dfrac{z-2}{3}$ along the line $\dfrac{x-9}{2}=\dfrac{y-13}{3}=\dfrac{z-17}{6}$ is
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The area (in sq. units) of the region A = {(x, y) : (x – 1)[x] $ \le $ y $ \le $ 2$\sqrt x $, 0 $ \le $ x $ \le $ 2}, where [t] denotes the greatest integer function, is :
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Solution
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