EQUATION OF A STRAIGHT LINE - SLOPE AND INTERCEPT

1. Find the slope of the following lines graphed using the points marked on the lines. Slope = (y2 – y1)/(x2 – x1) 1. Sample calculation: (4, -3) and (-5,4) are the two points on the line. Slope = [(4 -(-3)] Slope = 7/-9= -7/9 [(-5-4] Intercept is 0 2. Find the slope and intercept of each line on the Y axis a) 8x + 3y = -9 3y = -8x -9 Divide by 3 Y = (-8/3) x -3 Slope = -8/3 and intercept -3 b) 4x + 5y = -10 5y = -4x -10 Divide by 5 Y = (-4/5) x – 2 Slope = -4/5 and intercept is -2 Relationship between slopes of lines a) Parallel lines will have same slope b) Perpendicular lines will have the product of their slopes equal to -1 1. One line passes through the points (–1, –2) and (1, 2); another line passes through the points (–2, 0) and (0, 4). Are these lines parallel, perpendicular, or neither? Slope of one line = ( -2-2)/(-1-1) = -4/-2= 2 Slope of the other line = (4-0)/[0-(-2)]=4/2 = 2 The lines are parallel 2. One line passes through the points (0, –4) and (– 1, –7); another line passes through the points (3, 0) and (–3, 2). Are these lines parallel, perpendicular, or neither? Slope of one line = (-4 +7)/(0+1)=3 Slope of another line =(2-0)/(-3-3)=2/-6=-1/3 The lines are perpendicular Equation of a straight line 1. y = mx +b m is the slope and b is the intercept on y axis Find the equation of a line with slope 5 passing through the point (2,0). Y = 5x + b Find b by plugging in the values of x and y 0 = 5*2 + b b = -10 y= 5x -10 is the equation of the line 2. Find the equation of a line passing through (-4,2) and (1,3). Slope = (3-2)/(1+4) = 1/5 Y =mx +b y=(1/5)x + b The point(-4,2) lies on the line and hence should satisfy the equation. Y=(1/5)x + b 2= (1/5)(-4) + b b=2-4/5 b= 6/5 y= (1/5) x + 6/5 or 5y-x-6=0 is the equation of the line