How do you write an equation of a line passing through (2, 4), perpendicular to #10x - 5y = 8#?

2 Answers
Nam D.
Jan 8, 2018

#y=-1/2x+5#

Explanation:

First, convert #10x-5y=8# into the form of #y=mx+b#

#10x-5y=8#

#-5y=8-10x#

#y=2x-8/5#

The perpendicular line to #y=2x-8/5# will have a slope equal to #-1/2#, because perpendicular lines' slopes have a product of #-1#.

The new line will be

#y=-1/2x+b#

We know that it passes through #(2,4)#, so we can plug that into the new equation.

#4=-1/2*2+b#

#4=-2+b#

#b=6#

So, the new equation of the line will be

#y=-1/2x+5#

Jim G.
Jan 8, 2018

#y=-1/2x+5#

Explanation:

#"given a line with slope m then the slope of a line "#
#"perpendicular to it is"#

#•color(white)(x)m_(color(red)"perpendicular")=-1/m#

#"the equation of a line in "color(blue)"slope-intercept form"# is.

#•color(white)(x)y=mx+b#

#"where m is the slope and b the y-intercept"#

#"rearrange "10x-5y=8" into this form"#

#rArr-5y=-10x+8rArry=2x-8/5rArrm=2#

#rArrm_(color(red)"perpendicular")=-1/2#

#rArry=-1/2x+blarrcolor(blue)"is the partial equation"#

#"to find b substitute "(2,4)" into the partial equation"#

#4=-1+brArrb=4+1=5#

#rArry=-1/2x+5larrcolor(red)"perpendicular equation"#

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