How do you write an equation in point-slope form for the given y-intercept (0, 5) and is perpendicular to the line with equation y = –3x + 1?

Question

Avery Alexander · Accepted Answer

#y = 1/3x + 5#

Given Line
Red line represents this line - y = -3 +1

Blue line is the required line. This line is vertical to y = -3 +1 and passing through the point (0, 5)

Since the two lines are vertical, the product of their slopes is equal to - 1

#m_1 xx m_2 = - 1#
#m_1# is the slope of the given line(Red line)
#m_2# is the slope of the required line (blue)

# -3 xx m_2 = -1#
# m_2 = (-1)/(-3) = 1/3#

#y - y_1 = m_2(x - x_2)#
#y - 5 = 1/3 (x- 0)#

The equation of the line is -
#y = 1/3x + 5#

Jace Anderson · Answer

undefined To write an equation in point-slope form for a line perpendicular to the given line with equation $ y = -3x + 1 $ and passing through the point (0, 5), first find the slope of the given line. The slope of the given line is -3.

Since the line we want to find is perpendicular to the given line, its slope will be the negative reciprocal of -3, which is $ \frac{1}{3} $.

Now, we have the slope $ m = \frac{1}{3} $ and the point $ (0, 5) $. Using the point-slope form equation $ y - y_1 = m(x - x_1) $, we substitute $ x_1 = 0 $, $ y_1 = 5 $, and $ m = \frac{1}{3} $ to get the equation:

$$ y - 5 = \frac{1}{3}(x - 0) $$

Simplify to obtain the equation in point-slope form:

$$ y - 5 = \frac{1}{3}x $$