# Solutions to Exercises on Slope

The solutions to the exercises in the tutorial on the slope of a line are presented here.

 Matched Exercise to Example 1: Find the slope of the line passing through the pairs of points and describe the line as rising, falling, horizontal or vertical. (3 , -1) , (3 , 5) (-1 , 0) , (3 , 7) (2 , 1) , (6 , 0) (-5 , 2) , (9 , 2) Solution: The slope of the line is given by m = ( 5 - (-1) ) / (3 - 3) The denominator is equal to zero and therefore the slope is undefined. The slope of the line is given by m = ( 7 - 0 ) / ( 3 - (-1) ) = 7 / 4 Since the slope is positive, the line rises as x increases. We first find the slope of the line m = ( 0 - 1 ) / ( 6 - 2 ) = -1/4 Since the slope is neagtive, the line falls as x increases. The slope of the line is given by m = ( 2 - 2 ) / ( 9 - (-5) ) = 0 The slope is equal to zero, the line is horizontal. Matched Exercise to Example 2: A line has a slope of 5 and passes through the point (1 , -4). Find another point A through which the line passes. (many possible answers)Solution Let x1 and y1 be the x and y coordinates of point A. According to the definition of the slope ( y1 - (-4) ) / (x1 - 1) = 5 We need to solve this equation in order to find x1 and y1. This equation has two unknowns and therefore has an infinite number of pairs of solutions. We chose y1 and then find x1. if y1 = 11, for example, the above equation becomes ( 11 - (-4) ) / (x1 - 1) = 5 We obtain an equation in x1 15 / (x1 - 1) = 5 Solve for x1 to obtain x1 = 4 One possible answer is point A at (4 , 11) Check that the two points give a slope of 5 (11 - (-4) ) / (4 - 1) = 5 Matched Exercise to Example 3: Are the lines L1 and L2 passing through the given pairs of points parallel, perpendicular or neither parallel nor perpendicular? L1: (1 , 2) , (1 , 1) L2: (-4 , -1) , (-4 , 0) L1: (2 , 3) , (3 , 1) L2: (1 , -2) , (7 , -5) L1: (1 , -1) , (2 , -2) L2: (0 , 0) , (1 , 1) L1: (1 , 9) , (-2 , 9) L2: (18 , -1) , (0 , -1) SolutionIn what follows, m1 is the slope of line L1 and m2 is the slope of line L2. m1 = ( 1 - 2 ) / ( 1 - 1) m2 = ( 0 - (-1) ) / ( -4 - (-4) ) The two slopes m1 and m2 are undefined since the denominators in m1 and m2 are both equal to zero. Hence the two lines are vertical and therefore parallel m1 = ( 1 - 3 ) / ( 3 - 2 ) = -2 m2 = ( -5 - (-2) ) / ( 7 - 1 ) = -1/2 The product of the two slopes m1*m2 is not equal to -1, the two lines are neither perpendicular nor parallel. m1 = ( -2 - (-1) ) / ( 2 - 1 ) = -1 m2 = ( 1 - 0 ) / ( 1 - 0 ) = 1 The product of m1 and m2 is equal to -1, hence the two line are perpendicular. m1 = ( 9 - 9 ) / ( -2 - 1 ) = 0 m2 = ( -1 - (-1) ) / ( 0 - 18 ) = 0 The two slopes are equal , the two lines are parallel. Also the two lines are horizontal lines. Matched Exercise to Example 4: Is it possible for two lines with positive slopes to be perpendicular?Solution No. If both slopes are positive, their product can never be equal to -1.