Question

What is the equation of the line that is perpendicular to the given line and passes through the point (3, 4)?

y = –x 1/3+ 5
y = –x 1/3+ 3
y = 3x + 2
y = 3x − 5

Answer 1

Answer:

y = 3x - 5

Step-by-step explanation:

$\text{The slope-intercept form of an equation of a line:}\\y=mx+b\\m-slope\\b-y\ intercept$

$\text{The formula of a slope:}\\\\m=\dfrac{y_2-y_1}{x_2-x_1}$

$\text{Let}\ k:y=m_1x+b_1,\ l:y=m_2x+b_2,\ \text{then}\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2\\=========================$

$\text{From the graph we have two points (-3, 2) and (0, 1).}\\\text{Calculate the slope of the given line:}\\\\m=\dfrac{1-2}{0-(-3)}=\dfrac{-1}{3}=-\dfrac{1}{3}.\\\\\text{Therefore the slope of the perpendicular line is:}\\\\m=-\dfrac{1}{-\frac{1}{3}}=3\\\\\text{Put it and the coordinates of the point (3, 4) to the equation of a line:}\\\\4=3(3)+b\\4=9+b\qquad\text{subtract 9 from both sides}\\-5=b\to b=-5\\\\\text{Finally:}\\\\y=3x-5$