Question bank/JEE Main/Mathematics

Let ABC be a triangle such that the coordinates of the vertex A
are (-3,1). Equation of the median through B is 2x + y -3 =0 and
equation of the angular bisector of C is $7x - 4y - 1 = 0$ Find the
slope of line BC.
ekuk ABC,d f=Hkqt gS ftl A (-3,1) gSA B ls
tkus okyh lehdj.k 2x + y - 3 =0 C s.k vnZ~ lehdj.k $7x - 4y - 1 = 0$gS rc js[kk BC

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Since C lies on 7x - 4y - 1=0, therefore let us choose its
coordinates as $\left( h,\frac{7h - 1}{4} \right)$
[IMAGE] The mid point of AC, i.e.
$\left( \frac{h - 3}{2},\frac{7h + 3}{8} \right)$ lies on 2x + y -3 =0
therefore we have
$\left( \frac{h - 3}{2} \right) + \left( \frac{7h + 3}{8} \right) - 3 = 0$
gives h = 3
Hence, coordinates of C are (3, 5) and equation of AC is
$y - 5 = \frac{5 - 1}{3 + 3}(x - 3)$ i.e., $2x - 3y + 9 = 0$........
(i)
Let slope of BC = m. Since lines BC and AC
$\left( \ slope\ = \frac{2}{3} \right)$ are equally inclined to the
line
7x- 4y-1 =0 $\left( \ slope\ = \frac{7}{4} \right)$ therefore we have
i.e.,
$\frac{m - \frac{7}{4}}{1 + \frac{7m}{4}} = \frac{\frac{7}{4} - \frac{2}{3}}{1 + \frac{7}{6}}$
(see figure)
i.e., $\frac{4m - 7}{7m + 4} = \frac{1}{2}$gives m=

Source: Grass-G / M-26 Mathematics Paper 1 15 Jan.docx