Solution to Problem 1

Substituting the coordinates of the given points A , B and (x,y) in the equation a x + b y + c =0, we obtain the following linear system:

$\left\{ \begin{array}{r} \par a x + b y + c =0 \\ 3a +5b + c =0 \\ -1a + 2b + c =0 \\ \end{array} \right.$

With the matrix equation AX =B where A is the coefficient matrix $\left[ \begin{tabular}{rrr} x & y & 1 \\ 3 & 5 & 1 \\ -1 & 2 & 1 \end{tabular} \right ]$, $X = \left[ \begin{tabular}{r} a\\ b\\ c \end{tabular}\right]$and $B = \left[ \begin{array}{r} 0\\ 0\\ 0 \end{array}\right]$

Note that $\large AX = 0$ has a nonzero solution if and only if $\det(A) =0.$

Use cofactor expansion along the first line to find det(A).

$\det(A) = x \left\vert \begin{array}{rr} 5 & 1\\ 2 & 1\\ \end{array}\ri... ...+1 \left\vert \begin{array}{rr} 3 & 5\\ -1 & 2 \end{array}\right\vert = 0$

or

$\Large 4x -4y +11 = 0$. That is a= 4, b= -4 and c= 11.