Finding the slope of horizontal and vertical lines calculator

Example One

The slope of a line going through the point (1, 2) and the point (4, 3) is $$ \frac{1}{3}$$.

Remember: difference in the y values goes in the numerator of formula, and the difference in the x values goes in denominator of the formula.

The work , side by side

point (4, 3) as $$ (x_1, y_1 )$$

$$ slope = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{3-2}{4-1} = \frac{1}{3} $$

point (1, 2) as $$ (x_1, y_1 )$$

$$ slope = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{2-3}{1-4} = \frac{-1}{-3} = \frac{1}{3} $$

Answer: It does not matter which point you put first. You can start with (4, 3) or with (1, 2) and, either way, you end with the exact same number! $$ \frac{1}{3} $$

The slope of a vertical line is undefined

This is because any vertical line has a $$\Delta x$$ or "run" of zero. Whenever zero is the denominator of the fraction in this case of the fraction representing the slope of a line, the fraction is undefined. The picture below shows a vertical line (x = 1).

The slope of a horizontal line is zero

This is because any horizontal line has a $$\Delta y$$ or "rise" of zero. Therefore, regardless of what the run is (provided its' not also zero!), the fraction representing slope has a zero in its numerator. Therefore, the slope must evaluate to zero. Below is a picture of a horizontal line -- you can see that it does not have any 'rise' to it.

Problem 1

What is the slope of a line that goes through the points (10,3) and (7, 9)?

$ \frac{rise}{run}= \frac{y_{2}-y_{1}}{x_{2}-x_{1}} $

Using $$ \red{ (10,3)}$$ as $$x_1, y_1$$

$ \frac{9- \red 3}{7- \red{10}} \\ = \frac{6}{-3} \\ = \boxed {-2 } $

Using $$ \red{ (7,9)} $$ as $$x_1, y_1$$

$ \frac{3- \red 9}{10- \red 7} \\ =\frac{-6}{3} \\ = \boxed{-2 } $

Problem 2

A line passes through (4, -2) and (4, 3). What is its slope?

$ \frac{rise}{run}= \frac{y_{2}-y_{1}}{x_{2}-x_{1}} $

Using $$ \red{ ( 4,3 )}$$ as $$x_1, y_1$$

$ = \frac{-2 - \red 3}{4- \red 4} \\ = \frac{-5}{ \color{red}{0}} \\ = \text{undefined} $

Using $$ \red{ ( 4, -2 )}$$ as $$x_1, y_1$$

$ = \frac{3- \red{-2}}{4- \red 4} \\ = \frac{5}{ \color{red}{0}} \\ = \text{undefined} $

Whenever the run of a line is zero, the slope is undefined. This is because there is a zero in the denominator of the slope! Any the slope of any vertical line is undefined .

Problem 3

A line passes through (2, 10) and (8, 7). What is its slope?

$ \frac{rise}{run}= \frac{y_{2}-y_{1}}{x_{2}-x_{1}} $

Using $$ \red{ ( 8, 7 )}$$ as $$x_1, y_1$$

$ \frac{10 - \red 7}{2 - \red 8} \\ = \frac{3}{-6} \\ = -\frac{1}{2} $

Using $$ \red{ ( 2,10 )}$$ as $$x_1, y_1$$

$ \frac{7 - \red {10}}{8- \red 2} \\ = \frac{-3}{6} \\ = -\frac{1}{2} $

Problem 4

A line passes through (7, 3) and (8, 5). What is its slope?

$ \frac{rise}{run}= \frac{y_{2}-y_{1}}{x_{2}-x_{1}} $

Using $$ \red{ (7,3 )}$$ as $$x_1, y_1$$

$$ \frac{ 5- \red 3}{8- \red 7} \\ = \frac{2}{1} \\ = 2 $$

Using $$ \red{ ( 8,5 )}$$ as $$x_1, y_1$$

$$ \frac{ 3- \red 5}{7- \red 8} \\= \frac{-2}{-1} \\ = 2 $$

Problem 5

A line passes through (12, 11) and (9, 5) . What is its slope?

$ \frac{rise}{run}= \frac{y_{2}-y_{1}}{x_{2}-x_{1}} $

Using $$ \red{ ( 5, 9)}$$ as $$x_1, y_1$$

$$ \frac{ 11 - \red 5}{12- \red 9} \\ = \frac{6}{3} \\ =2 $$

Using $$ \red{ (12, 11 )}$$ as $$x_1, y_1$$

$$ \frac{ 5- \red{ 11} }{9- \red { 12}} \\ = \frac{-6}{-3} \\ = 2 $$

Problem 6

What is the slope of a line that goes through (4, 2) and (4, 5)?

$ \frac{rise}{run}= \frac{y_{2}-y_{1}}{x_{2}-x_{1}} $

Using $$ \red{ ( 4,5 )}$$ as $$x_1, y_1$$

$$ \frac{ 2 - \red 5}{4- \red 4} \\ = \frac{ -3}{\color{red}{0}} \\ = undefined $$

Using $$ \red{ ( 4,2 )}$$ as $$x_1, y_1$$

$$ \frac{ 5 - \red 2}{4- \red 4} \\ = \frac{ 3}{\color{red}{0}} \\ = undefined $$

Challenge Problem

Find the slope of A line Given Two Points.

Attempt #1

$ slope= \frac{rise}{run} \\= \frac{\color{red}{y_{2}-y_{1}}}{\color{blue}{x_{2}-x_{1}}} \\= \frac{6-3}{1-2} \\= \frac{3}{-1} =\boxed{-3} $

Attempt #2

$$ slope= \frac{rise}{run} \\= \frac{\color{red}{y_{2}-y_{1}}}{\color{blue}{x_{2}-x_{1}}} \\= \frac{6-3}{2-1} \\= \frac{3}{1} \\ = \boxed{3} $$

Attempt #3

$$ slope = \frac{rise}{run} \\= \frac{\color{red}{y_{2}-y_{1}}}{\color{blue}{x_{2}-x_{1}}} \\ =\frac{2-1}{6-3} \\ =\boxed{ \frac{1}{3}} $$

The correct answer is attempt #2.

In attempt #1, she did not consistently use the points. What she did, in attempt one, was :

$$ \frac{\color{red}{y{\boxed{_2}}-y_{1}}}{\color{blue}{x\boxed{_{1}}-x_{2}}} $$

The problem with attempt #3 was reversing the rise and run. She put the x values in the numerator( top) and the y values in the denominator which, of course, is the opposite!

$$ \cancel {\frac{\color{blue}{x_{2}-x_{1}}}{\color{red}{y_{2}-y_{1}}}} $$