Question:
Ask and be perpendicular to the straight line of $y = x -1$ and lead the straight line equation that cross a point {4,3}
.
My method is:
{eqn = y == x - 1, pt = {4, 3}}
eq = # == 0 &@(eq /. Equal -> Subtract)
line = Cross@(Coefficient[eq[[1]], #] & /@ {x, y}) . {x, y} + c == 0
sol = First@(line /. Thread[{x, y} -> pt] // Solve)
line = line /. sol // Simplify // SubtractSides
This point may not necessarily lie on the original line. For example, a line that is perpendicular to the line y == x - 1
and passes through the point {6, 9}
.
-15 + x + y == 0
Are there other methods to solve this problem?