1st box: 2a (subtract 2a from both sides)
2nd box: 3 (add 3 to both sides)
3rd box: 10 (add 7+3)
4th and 5th box: 2 (divide both by 2)
6th box: 5 (10/2=5)
As We can see that point J is on (-2,5) and L is on (-2,-2)
So we know that distance cannot be in negative, thus
Distance between J and L =5+2=7 units
JL=7
1.Write an equation in slope- intercept form of the line that passes through the given point and is parallel to the graph of the given equation. (2,-2);y=-x-2
D.y=-x
2.Write an equation in slope- intercept form of the line that passes through the given point and is parallel to the graph of the given equation. (2,-1);y=-3/2x-6
C.y=-3/2x+2
3.Write an equation in slope- intercept form of the line that passes through the given point and is parallel to the graph of the given equation. (4,2);x=-3
D.y=4
4.Write an equation in slope- intercept form of the line that passes through the given point and is perpendicular to the graph of the given equation. (-2,3);y=1/2x-1
B.y=-2x-1
5.Write an equation in slope- intercept form of the line that passes through the given point and is perpendicular to the graph of the given equation. (5,0);y+1=2(x-3)
D.y=-1/2x+5/2
