Question bank/JEE Main/Mathematics

Let $A \equiv (0,0),B \equiv (4,4)$ and $C \equiv (8,2)$ be the
vertices of a $\mathrm{\Delta}ABC$. If straight lines through the points
B and C meet at point P such that area of
$\mathrm{\Delta}PBC = \frac{1}{3}$area of $\mathrm{\Delta}ABC\$, then
locus of P can be
ekuk $A \equiv (0,0),B \equiv (4,4)$ rFkk $C \equiv (8,2)$f=kHkqt
rFkk C ls xqtjrh gS rFkk fcUnq P ij bl izdkj feyrh gS fd
$\mathrm{\Delta}ABC\$ =Qy] rc P

A$x - 4y + 4 = 0$B$x + 2y = 32$C$x + 2y = 8$D$x - y - 2 = 0$

Show answer and solution▾

[IMAGE] Required locus must be a straight line passing through centroid of the
$\mathrm{\Delta}ABC$ and parallel to side BC
$$\therefore\ Centroid\ \equiv (4,2)$$
vHkh"V fcUnqiFk,d ljy js[kk gksxh tks f=Hkqt $\mathrm{\Delta}ABC$
Unzd ls gksdj xqtjrh gS rFkk js[kk BC

Source: Grass-G / M-25 Mathematics Paper 7 Jan.docx