How do you write and equation of a line given (10, –2) with a slope of –1?

2 Answers
Oct 24, 2017

#y=-x+8# or graph{y=-x+8 [-22.8, 22.83, -11.4, 11.4]}

Explanation:

The equation of a line is #y=mx+b#.

Since we know that #m=-1# (#m# is the slope), the equation will be #y=(-1)x+b#

Put in (10, -2) as #x=10# and #y=-2#:

#-2=(-1)*10+b->#
#-2=-10+b-># (Multiply #-1# and #10#)
#8=b# (Add #10# to both sides)
Flip equation: #b=8# (Switch sides)

Whole equation: #y=-x+8#

#-x# because #-1*x=-x#

Ans: #y=-x+8#

Hopefully I'm right ;/

Oct 24, 2017

#y=-x+8#

Explanation:

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

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=mx+b)color(white)(2/2)|)))#

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

#rArry=-x+blarr" is the partial equation"#

#"to find b substitute "(10,-2)" into the partial equation"#

#-2=-10+brArrb=8#

#rArry=-x+8larrcolor(red)" in slope-intercept form"#
graph{-x+8 [-10, 10, -5, 5]}