How do you find a standard form equation for the line with #(7,-4)# and perpendicular to the line whose equation is #x-7y-4=0#?

2 Answers
Oct 5, 2017

Answer:

#7x+y=45#

Explanation:

#"the equation of a line in "color(blue)"standard form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(Ax+By=C)color(white)(2/2)|)))#
where A is a positive integer and B, C are integers.

#• " 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#

#"rearrange "x-7y-4=0" into "color(blue)"slope-intercept form"#

#•color(white)(x)y=mx+b#

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

#rArry=1/7x-4/7rArrm=1/7#

#rArrm_(color(red)"perpendicular")=-1/(1/7)=-7#

#rArry=-7x+blarr" partial equation"#

#"to find b substitute "(7,-4)" into the partial equation"#

#-4=-49+brArrb=45#

#rArry=-7x+45larrcolor(red)" in slope-intercept form"#

#rArr7x+y=45larrcolor(red)" in standard form"#

Oct 5, 2017

Answer:

#7x+y=45#

Explanation:

The given equation is in standard form which is #ax +by=c#

but we need to know its slope, so change it into the form #y=mx+c#

#x-7y-4 =0" "rarr x-4 =7y" "rarr7y = x-4 #

#y=color(blue)(1/7)x -4/7#

#rarrm=1/7#

If lines are perpendicular then #m_1 xx m_2 = -1#

(One slope is the negative reciprocal of the other - flip and change the sign.)

The slope perpendicular to #1/7# is #-7#

Now we have a point #(7,-4)# and #m =-7# so substitute into the point-slope formula for a straight line.

#y-y_1 =m(x-x_1)#

#y-(-4) = -7(x-7)#

#y+4 = -7x+49" "larr# change to standard form:

#7x+y = 49-4#

#7x+y=45#