We have,
Since, G(2,a), so
31+α+β+3=2 and 3−2−2α+β=a
⇒α+β=2 and −2α+β=3a+2
So, α=−a,β=2+a
So, B≡(−a,2a),C≡(5+a,2+a)
Now, B and C lies on the line x+py=21a, so
−a+2pa=21a
⇒p=11
Also,
5+a+11(2+a)=21a
⇒27=9a⇒a=3
So,
B≡(−3,6),C≡(8,5)
So,
BC2=121+1=122
The equations of the sides AB,BC&CA of a triangle ABC are 2x+y=0, x+py=21a (a=0) and x−y=3 respectively. Let P(2,a) be the centroid of the triangle ABC, then (BC)2 is equal to
Held on 24 Jan 2023 · Verified 6 Jul 2026.
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.