How do you write an equation of a line passing through (0, -5), perpendicular to #4x+y=0 #?

Question

How do you write an equation of a line passing through (0, -5),  perpendicular to #4x+y=0 #?

Julia Adams · Accepted Answer

#x-4y=20# For the general case: #color(white)("XXX")Ax+By=C# has a slope of #m=-A/B# So #4x+1=0# has a slope of #m=-4# In general if a line has a slope of #m# all lines perpendicular to it will have a slope of #-1/m# There any line perpendicular to #4x+y=0# will have a slope of #1/4#. If such a perpendicular line passes through #(0,-5)# then using the slope-point form: #color(white)("XXX")y-(-5)=1/4(x-0)# or #color(white)("XXX")x-4y=20#

Isaiah Anthony · Answer

undefined To write an equation of a line passing through the point (0, -5) and perpendicular to the line 4x + y = 0, first find the slope of the given line.

The given line's slope can be found by rearranging the equation into slope-intercept form $ y = mx + b $, where $ m $ is the slope.

$ 4x + y = 0 $ becomes $ y = -4x $, so the slope of the given line is $ m = -4 $.

Since the line we're looking for is perpendicular to the given line, its slope will be the negative reciprocal of the given line's slope. So, the slope of the new line will be $ m = \frac{-1}{-4} = \frac{1}{4} $.

Now, using the point-slope form of the equation of a line $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) $ is the given point and $ m $ is the slope, we can plug in the values:

$ y - (-5) = \frac{1}{4}(x - 0) $

$ y + 5 = \frac{1}{4}x $

Now, we can rearrange the equation to the standard form:

$ y = \frac{1}{4}x - 5 $