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

Dear Student,

we are given with the three equations of lines of the triangle,

$$\begin{gathered} 4x-7y=-10 -\left(1\right) line AC \\ 7x+4y=15 -\left(2\right) line AB \\ x+y=5 -\left(3\right) line BC \\ solving e{q}^n 1 and 2 we get coordinates of point A(1,2) \\ solving e{q}^n 2 and 3 we get coordinates of point B(\frac{5}{3},\frac{20}{3}) \\ solving e{q}^n 1 and 3 we get coordinates of point C(\frac{25}{11},\frac{30}{11}) \end{gathered}$$
slope of line BC= 
$$\begin{gathered} y=mx+c \\ y=-x+5, hence m=-1 \end{gathered}$$

as we know the slope of perpendicular to line BC will be 
$$\begin{gathered} m_1\times m_2=-1 \\ putting m_1=-1 \\ {m}_2=1 \end{gathered}$$                                                    

equation of perpendicular from A to line BC

$$\begin{gathered} y-y_1=m(x-x_1) \\ y-2=1(x-1) \\ y=x+1 \end{gathered}$$

similarly the slope of line AC


$$\begin{gathered} 7y=4x+10 \\ y=\frac{4}{7}x+\frac{10}{7} \\ slope =\frac{4}{7} \end{gathered}$$

slope of perpendicular from B to AC 
$$\begin{gathered} -\frac{7}{4} \\ equation of perpendicular from B to AC \\ y-\frac{20}{3}=-\frac{7}{4}(x-\frac{5}{3}) \\ 21x+12y=115 \end{gathered}$$

solving both equations
$$\begin{gathered} y=x+1 \\ 21x+12y=115 \\ we get \left(\frac{103}{9},\frac{112}{9}\right) as orthocentre \end{gathered}$$

Regards