Easy way to find coordinates of intersection heights using vectors?

Question

What is the most simple way for me to find coordinates of intersection of heights if I'm given coordinates for points of triangle A(3,4), B(7,6), C(1,3) for example in cartesian plane? I know how to do it with linear equations, but I'm not sure if I can do it with vectors.

Answer 1

Let $H$ (orthocentre) denote the common point to all altitudes.

The different orthogonality conditions are written as dot products equal to 0:

$$\tag{1}\cases{\vec{HA}\cdot\vec{BC}=0\\\vec{HB}\cdot\vec{CA}=0}$$

(see remark below) which can be written under the form:

$$\tag{2} \cases{(x_A-x)(x_C-x_B)+(y_A-y)(y_C-y_B)=0 \ \ (a) \\(x_B-x)(x_A-x_C)+(y_B-y)(y_A-y_C)=0 \ \ (b)}$$

which is a system of two linear equations in the two unknowns $x,y.$

I let you do the final numerical computation.

Remark 1 : in (1), we had no need for the third condition $\vec{HC}\cdot\vec{AB}=0$, which is a consequence of the others.

Remark 2 : (2)(a) is nothing else than the equation of the altitude through point $A$ which is orthogonal to line $BC$. The same for (2) (b).