-x+y=0 and y=x+3
you want the perpendicular distance between the two lines
find the equation of one perpendicular line between the two parallel lines (they are parallel because they have the same slope)
y=x+3
y=x
draw the perpendicular line from the point (0,3) on line y=x+3
the slope of the perpendicular line is the negative reciprocal of one of the parallel lines, that is -1
the y-intercept is 3
the equation of the line is y=-x+3
using one of the parallel lines, y=x, and the perpendicular line, y=-x+3, you have a system of linear equations and you solve the system for (x,y), the point of intersection of the lines and therefore the other endpoint of the perpendicular line
y=x
y=-x+3
substitute
x=-x+3
x+x=3
2x=3
x=3/2
therefore y=3/2 also
now you have the endpoints of the perpendicular distance between the parallel lines
use the distance formula with...
(0,3) and (3/2,3/2)
√[(3-1.5)^2+(0-1.5)^2]
√[(1.5)^2+(1.5)^2]
√[(3/2)^2+(3/2)^2]
√[(9/4)+(9/4)]
√(18/4)
√18/√4
3√2/2 is the perpendicular distance between the two parallel lines
another solution...
because one line is y=x, draw a line from (0,3) to (3,0)
this forms a right isosceles triangle
using the Pythagorean Theorem...
3^2+3^2=P^2
9+9=P^2
18=P^2
P=√18
P=3√2, the hypotenuse of the triangle and the line that you drew
this line is half of the perpendicular distance between the parallel line
thus, 3√2/2 is the perpendicular distance between the two parallel lines
