The orthocentre of the $\Delta$ formed by the lines $x+y=1$, $2 x+3 y=6,4 x-y+4=0$ lies in the quadr

Question

The orthocentre of the $\Delta$ formed by the lines $x+y=1$, $2 x+3 y=6,4 x-y+4=0$ lies in the quadrant no. .........

Professor PC Thomas Classes · Answer

Solving in pairs, the vertices are\ $A\left(-\frac{3}{7}, \frac{16}{7}ight)$,\ $B\left(-\frac{3}{5}, \frac{8}{5}ight)$\ and $C(-3,4)$\ Let $H(h, k)$ be the orthocentre of $	riangle A B C$.\ (Slope of CH) (Slope of $A B$ ) $=-1$\ gives $h+4 k=13$\ (Slope of $A H$ ) (Slope of $B C$ ) $=-1$ gives $h-k$ $=\frac{-19}{7}$\ $	herefore h=\frac{3}{7}, k=\frac{22}{7} \quad 	herefore H=\left(\frac{3}{7}, \frac{22}{7}ight)$\ $	herefore \quad H$ lies in the first quadrant\