Algebra 1

line a: $y = 3x + 3$ line b: $x = -1$ line c: $y - 5 = 1/2* (x-2)$ line d: $y = 3$ line e: $y + 4 = -2*(x+6)$ line f: $9x-3y =5$ Converting all lines (where possible) to slope-intercept form lines a, b, and d are already in slope-intercept form (or are horizontal or vertical lines) C: $y - 5 = 1/2* (x-2)$ $y - 5 = 1/2*x -1$ $y - 5+5 = 1/2*x -1+5$ $y = 1/2*x +4$ E: $y + 4 = -2*(x+6)$ $y + 4 = -2x - 12$ $y + 4 - 4 = -2x - 12 -4$ $y = -2x - 16$ F: $9x - 3y = 5$ $9x - 3y +3y -5= 5+3y -5$ $9x - 5 = 3y$ $3y = 9x-5$ $3y/3 = (9x-5)/3$ $y = 3x - 5/3$ line a: $y = 3x + 3$ line b: $x = -1$ line c: $y = 1/2*x +4$ line d: $y = 3$ line e: $y = -2x - 16$ line f: $y = 3x - 5/3$ Parallel lines have the same slope. Thus lines a and f are parallel ( $m= 3$). Perpendicular lines have a product of their slopes equal to -1. Thus, lines c and e are perpendicular. Also, lines b and d are perpendicular since the angle made by the two lines is a 90 degree angle (definition of perpendicular).