Question

What is the equation of the line that passes through the point (3, 0) and is perpendicular to the line 2x – y = 5?

Answer 1

Answer:

$\large\boxed{y=-\dfrac{1}{2}x+\dfrac{3}{2}}$

Step-by-step explanation:

$\text{Let}\ k:y=_1x+b_1\ \text{and}\ l:y=m_2x+b_2.\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\============================\\\\\text{We have}\ 2x-y=5.\\\\\text{Convert to the slope-intercept form y = mx + b:}\\\\2x-y=5\qquad\text{subtract 2x from both sides}\\\\-y=-2x+5\qquad\text{change the signs}\\\\y=2x-5\to m_1=2\\\\\text{Therefore}\ m_2=-\dfrac{1}{2}.$

$\text{The equation of the searched line:}\ y=-\dfrac{1}{2}x+b.\\\\\text{The line passes through }(3,\ 0).\\\\\text{Put the coordinates of the point to the equation.}\ x=3,\ y=0:\\\\0=-\dfrac{1}{2}(3)+b\\\\0=-\dfrac{3}{2}+b\qquad\text{add}\ \dfrac{3}{2} \text{to both sides}\\\\b=\dfrac{3}{2}$