Question

Write an equation that is perpendicular to the line y = 2x + 13?

Answer 1

$k:\ y=m_1x+b_1\ \ \ and\ \ \ l:\ y=m_2x+b_2\\\\k\perp l\ \ \ \Leftrightarrow\ \ \ m_1\cdot m_2=-1\\------------------------\\\\k:\ y=2x+13\ \ \ \ and\ \ \ \ l:\ y=m_2x+b_2\\\\2\cdot m_2=-1\ \ \ \Leftrightarrow\ \ \ m_2=- \frac{1}{2} \\\\l:\ y=- \frac{1}{2} x+b_2\ \ \ and\ \ \ b_2\ \in\ R\\\\for\ example:\\\\y=- \frac{1}{2} x,\ \ \ y=- \frac{1}{2} x+5 \frac{1}{7} ,\ \ \ y=- \frac{1}{2}x - \sqrt{13}\ ,\ ...$

Answer 2

An equation can't be perpendicular to a line, but the graph of the equation can.

When that happens, we recall that the slopes of perpendicular lines are negative reciprocals.

One line is the graph of [ y = 2x + 13 ].  The slope of the line is 2 .
So the slope of a line perpendicular to it must be -1/2 .

The equation of a line perpendicular to it is:    y = (-1/2 x) plus (any number).

The y-intercept of the perpendicular line doesn't matter.  Only its slope does.