# Slope of Lines Questions with Solutions

Questions on the slope of lines and matched exercises with detailed solutions and explanations are presented.

## Questions with Solutions

Question 1
Find the slope of the line passing through the pairs of points and describe the line as rising, falling, horizontal or vertical.
1. (2 , 1) , (4 , 5)
2. (-1 , 0) , (3 , -5)
3. (2 , 1) , (-3 , 1)
4. (-1 , 2) , (-1 ,- 5)
Solution to Question 1
1. The slope of the line is given by
m = ( 5 - 1 ) / (4 - 2) = 4 / 2 = 2
Since the slope is positive, the line rises as x increases.
2. The slope of the line is given by
m = ( -5 - 0 ) / ( 3 - (-1) ) = -5 / 4
Since the slope is negative, the line falls as x increases.
3. We first find the slope of the line
m = ( 1 - 1 ) / ( -3 - 2 ) = 0
Since the slope is equal to zero, the line is horizontal (parallel to the x axis).
4. The slope of the line is given by
m = ( -5 - 2 ) / ( -1 - (-1) ) = - 7 / 0
Division by zero in not allowed in math. Therefore the slope of the line is undefined and the line is vertical. (parallel to the y axis).
Matched Exercise to Question 1 for more practice

Question 2
A line has a slope of - 2 and passes through the point (2 , 5). Find another point A through which the line passes. (many possible answers)
Solution to Question 2

• Let x1 and y1 be the x and y coordinates of point A. According to the definition of the slope
( y1 - 5 ) / (x1 - 2) = -2
• 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 x1 and then find y1. if x1 = -1, for example, the above equation becomes
( y1 - 5 ) / (- 1 - 2) = - 2
• We obtain an equation in y1
( y1 - 5 ) / -3 = - 2
• Solve for y1 to obtain
y1 = 11
• One possible answer is point A at
(- 1 , 11)
• Check that the two points give a slope of -2
(11 - 5 ) / (- 1 - 2) = 6 /-3 = - 2
Matched Exercise to Question 2 for more practice

Question 3
Are the lines L1 and L2 passing through the given pairs of points parallel, perpendicular or neither parallel nor perpendicular?

1. L1: (1 , 2) , (3 , 1)
L2: (0 , -1) , (2 , 0)
2. L1: (0 , 3) , (3 , 1)
L2: (-1 , 4) , (-7 , -5)
3. L1: (2 , -1) , (5 , -7)
L2: (0 , 0) , (-1 , 2)
4. L1: (1 , 0) , (2 , 0)
L2: (5 , -5) , (-10 , -5)
5. L1: (-2 , 5) , (-2 , 7)
L2: (5 , 1) , (5 , 13)
Solution to Question 3
In what follows,  m1 is the slope of line L1 and m2 is the slope of line L2.
1. Find the slope m1 of line L1 and the slope m2 of line L2
m1 = ( 1 - 2 ) / ( 3 - 1 ) = -1 / 2
m2 = ( 0 - (-1) ) / ( 2 - 0 ) = 1/2
The two slopes m1 and m2 are not equal and their product m1�m2 is not equal to - 1. Hence the two lines are neither parallel nor perpendicular.
2. m1 = ( 1 - 3 ) / ( 3 - 0 ) = -2 / 3
m2 = ( -5 - 4 ) / ( -7 - (-1) ) = -9 / -6 = 3/2
The product of the two slopes m1 � m2 = (-2 / 3)(3 / 2) = -1, the two lines are perpendicular.
3. m1 = ( -7 - (-1) ) / ( 5 - 2 ) = -6 / 3 = -2
m2 = ( 2 - 0 ) / ( -1 - 0 ) = -2
The two slopes are equal, the two lines are parallel.
4. m1 = ( 0 - 0 ) / ( 2 - 1 ) = 0 / 1 = 0
m2 = ( -5 - (-5) ) / ( - 10 - 5 ) = 0 / -15 = 0
The two slopes are equal , the two lines are parallel. Also the two lines are horizontal since their slopes are equal to zero.
5. m1 = ( 7 - 5 ) / ( - 2 - (-2) ) = 2 / 0 = undefined
m2 = ( 13 - 1 ) / ( 5 - 5 ) = 12 / 0 = undefined
The two slopes are both undefined since the denominators in both m1 and m2 are equal to zero. The two lines are vertical and therefore parallel.
Matched Exercise to Question 3 for more practice

Question 4
Is it possible for two lines with negative slopes to be perpendicular?
Solution to Question 4

No. The product of the slopes of two perpendicular lines is equal to -1. If both slopes are negative, their product can never be equal to - 1.

Matched Exercise to Question 4 for more practice

## More matched questions with solutions

Matched Exercise to Question 1
Find the slope of the line passing through the pairs of points and describe the line as rising, falling, horizontal or vertical.

1. (3 , - 1) , (3 , 5)
2. (- 1 , 0) , (3 , 7)
3. (2 , 1) , (6 , 0)
4. (- 5 , 2) , (9 , 2)
Solution to Matched Exercise to Question 1.

Matched Exercise to Question 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 to Matched Exercise to Question 2.

Matched Exercise to Question 3
Are the lines L1 and L2 passing through the given pairs of points parallel, perpendicular or neither parallel nor perpendicular?

1. L1: (1 , 2) , (1 , 1)
L2: (-4 , -1) , (-4 , 0)
2. L1: (2 , 3) , (3 , 1)
L2: (1 , -2) , (7 , -5)
3. L1: (1 , -1) , (2 , -2)
L2: (0 , 0) , (1 , 1)
4. L1: (1 , 9) , (-2 , 9)
L2: (18 , -1) , (0 , -1)
Solution to Matched Exercise to Question 3.

Matched Exercise to Question 4
Is it possible for two lines with positive slopes to be perpendicular?
Solution to Matched Exercise to Question 4.

## Solution to Matched Exercise Above

Solution to Matched Exercise in Question 1

1. 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.

2. 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.

3. We first find the slope of the line
m = ( 0 - 1 ) / ( 6 - 2 ) = -1/4
Since the slope is negative, the line falls as x increases.

4. The slope of the line is given by
m = ( 2 - 2 ) / ( 9 - (-5) ) = 0
The slope is equal to zero, the line is horizontal.

Solution to Matched Exercise in Question 2
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

Solution to Matched Exercise in Question 3
In what follows,  m1 is the slope of line L1 and m2 is the slope of line L2.

1. m1 = ( 1 - 2 ) / ( 1 - 1) = - 1 / 0
2. m2 = ( 0 - (-1) ) / ( -4 - (-4) ) = 1 / 0
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

3. m1 = ( 1 - 3 ) / ( 3 - 2 ) = -2
4. m2 = ( -5 - (-2) ) / ( 7 - 1 ) = -1/2
The product m1�m2 of the two slopes is not equal to -1, the two lines are neither perpendicular nor parallel.
5. m1 = ( -2 - (-1) ) / ( 2 - 1 ) = -1
6. m2 = ( 1 - 0 ) / ( 1 - 0 ) = 1
The product of m1 and m2 is equal to - 1, hence the two line are perpendicular.
7. m1 = ( 9 - 9 ) / ( -2 - 1 ) = 0
8. m2 = ( -1 - (-1) ) / ( 0 - 18 ) = 0
The two slopes are equal , the two lines are parallel. Also the two lines are horizontal lines.

Solution to Matched Exercise in Question 4

No. The product of the slopes of two perpendicular lines is equal to -1. If both slopes are positive, their product can never be equal to -1.