Answer :
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.
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.
[tex]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}\ ,\ ...[/tex]
