In a triangle ABC,the vertex A is (1,1) and orthocenter is (2,4).If the sides AB and BC are members of the family of straight lines $ax+by+c=0$,where $a,b,c$ are in arithmatic progression.Then the coordinates of vertex C are (h,k).Find the value of 2h+12k.
My attempt:Since the coordinates of B are not given directly,I do not have any idea to solve this problem.Plus a,b,c in AP is confusing me.Can someone help me in this question?
AB and BC are members of the family of straight lines ax+by+c=0,where a,b,c are in arithmatic progression. Therefore we can write
AB: $a_1x+b_1y+c_1=0$ and because A(1,1) lies on this line $\Rightarrow a_1+b_1+c_1=0$. Because $a_1,b_1,c_1$ are in AP follows $a_1+a_1+d+a_1+2d=0\Rightarrow a_1+d_1=0\Rightarrow b_1=0$. Consequently, the vector $(a_1,b_1)=(a_1,0)\perp$ AB $\Rightarrow$ the vector $(0,a_1)$ is colinear with AB. But $CH\perp AB\Rightarrow (2-h,4-k)\perp (0,a_1)\Rightarrow (4-k)a_1=0\Rightarrow k=4$ because $a_1\neq 0$.
For BC we proceed the same way: BC: $a_2x+b_2y+c_2=0$ and because $C(h,k)$ lies on this line $\Rightarrow a_2h+b_2k+c_2=0\Leftrightarrow a_2h+4(a_2+d_2)+a_2+2d_2=0\Leftrightarrow a_2h+5a_2+6d_2=0 \,\,(*)$. Now we use that the vector $\vec {AH}\,(2-1,4-1)\perp BC$, i.e $(1,3)\perp (-b_2,a_2)=(-a_2-d_2,a_2)\Rightarrow -a_2-d_2+3a_2=0\Rightarrow d_2=2a_2$. Plug this in $(*)$ and get $a_2h+5a_2+12a_2=0\Rightarrow h=-17$, because $a_2\neq 0$.