Using parametric form through P(2,3): point at distance r=32 is (2+rcosθ, 3+rsinθ).
Substituting in x+y=6: r(cosθ+sinθ)=1, so cosθ+sinθ=23.
2sin(θ+4π)=23⇒sin(θ+4π)=23.
θ+4π=3π or 32π.
θ1=12π, θ2=125π.
θ1+θ2=12π+125π=2π.
Let the angles made with the positive x-axis by two straight lines drawn from the point P(2,3) and meeting the line x+y=6 at a distance 32 from the point P be θ1 and θ2. Then the value of (θ1+θ2) is:
Held on 24 Jan 2026 · Verified 6 Jul 2026.
3π
6π
2π
12π
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Let the image of parabola $x^{2}=4 y$, in the line $x-y=1$ be $(y+a)^{2}=b(x-c)$, $a, b, c \in \mathrm{~N}$. Then $a+b+c$ is equal to
The distance between the points (3, 4) and (6, 8) is:
If the chord joining the points $\mathrm{P}_{1}\left(x_{1}, y_{1}\right)$ and $\mathrm{P}_{2}\left(x_{2}, y_{2}\right)$ on the parabola $y^{2}=12 x$ subtends a right angle at the vertex of the parabola, then $x_{1} x_{2}-y_{1} y_{2}$ is equal to
The distance between the parallel lines 3x + 4y - 7 = 0 and 3x + 4y + 8 = 0 is:
Let a point $A$ lie between the parallel lines $L_{1}$ and $L_{2}$ such that its distances from $L_{1}$ and $L_{2}$ are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle $A B C$, where the points $B$ and C lie on the lines $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$, respectively, is :
Work through every JEE Main Coordinate Geometry PYQ, year by year.