Question 995643
{{{y= 4x + 4}}}
{{{y= 4x - 4}}}

{{{ graph( 600, 600, -10, 10, -10, 10, 4x + 4, 4x - 4) }}}

to find the distance between the pair of parallel lines first find perpendicular line and its intersection points with given parallel lines

given parallel lines have same slope and it is {{{m=4}}}

y-intercept of first line is at ({{{0}}},{{{4}}}) and

y-intercept of second line is at ({{{0}}},{{{-4}}})

so, the perpendicular line will have a slope negative reciprocal to {{{m=4}}} and it is {{{m[p]=-1/4}}}

use the slope and y-intercept of first line is at ({{{0}}},{{{4}}}) to find the equation of the line perpendicular to the line {{{y= 4x + 4}}}

The equation of the line that has that {{{m[p]=-1/4}}} and goes through point ({{{0}}},{{{4}}}) is:

{{{y-y[p]=m(x-x[p])}}}

{{{y-4=-(1/4)x}}}

{{{y=-(1/4)x+4}}}


so, the lines {{{y= 4x + 4}}} and {{{y=-(1/4)x+4}}} are  perpendicular to each other and intersect at ({{{0}}},{{{4}}})

now we need to find intersection point of the  lines {{{y= 4x-4}}} and {{{y=-(1/4)x+4}}}

equal right sides and solve for {{{x}}}

{{{ 4x-4=-(1/4)x+4}}}

{{{ 4x+(1/4)x=4+4}}}

{{{ 4x+x/4=8}}}

{{{ 4*4x+4x/4=8*4}}}

{{{ 16x+x=32}}}

{{{ 17x=32}}}

{{{ x=32/17}}}

{{{ x=1.882352941176471}}}

{{{ x=1.9}}}

with corresponding {{{y}}},

{{{y=4x-4}}}

{{{y=4*1.9-4=3.6}}}


so, the  lines {{{y= 4x-4}}} and {{{y=-(1/4)x+4}}} intersect at ({{{1.9}}},{{{3.6}}})


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(0,4,.12),circle(1.9,3.6,.12),
locate(0,4,p(0,4)),locate(1.9,3.6,p(1.9,3.6)),
 graph( 600, 600, -10, 10, -10, 10, 4x + 4, 4x - 4,-(1/4)x+4)) }}}


now, find the distance between points ({{{0}}},{{{4}}}) and ({{{1.9}}},{{{3.6}}}); that will be the distance between given parallel lines


{{{d^2=(x[1]-x[2])^2+(y[1]-y[2])^2}}}

{{{d^2=(0-1.9)^2+(4-3.6)^2}}}

{{{d^2=(-1.9)^2+(0.4)^2}}}

{{{d^2=3.61+0.16}}}

{{{d^2=3.77}}}

{{{d=1.94164878389476}}}

{{{d=1.94}}}


so, your answer is c){{{1.94}}}