Question

The orthocentre of the triangle formed by the lines x+y=1 , 2x+3y=6 and 4x-y+4=0 lies in quadrant number

• A. $1^{st}$
• B. $2^{nd}$
• C. $3^{rd}$
• D. $4^{th}$

Answer 1

The correct option is A $1^{st}$

Let find out the cordinates from the given line

On solving line 1 and 2 we get A$(\frac{-3}{7},\frac{16}{7})$

On solving line 2 and 3 we get B$(-\frac{3}{5},\frac{8}{5})$

and On solving 1 and 3 we get last point C(-3,4)

Let H bet the orthocenter H(h,k)

as HA$\perp$BC

(Slope of HA).(slope of BC)=-1

$\therefore \frac{k-\frac{16}{7}}{h+\frac{3}{7}}\times (-1)=-1$

$\to \frac{7k-16}{7h+3}=1$

$\to 7k-16=7h+3$

$\to 7h-7k+19=0\left(1\right)$

Also

HC$\perp$AB

(slope of HC).(slope of AB)=-1

$\therefore \frac{k-4}{h+3}\times 4=-1$

$\to 4k-16=-h-3$

$\to h+4k-13=0-\left(2\right)$

On solving eqs.(1) and (2) we get h=$\frac{3}{7},k=\frac{22}{7}$

So orthocenter lies in first quadrant.