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Negative Slope: Definition and Examples | EDU.COMEDU.COMResourcesBlogGuidePodcastPlanBackHomesvg]:size-3.5">Math Glossarysvg]:size-3.5">Negative SlopeNegative Slope: Definition and ExamplesTable of ContentsNegative Slope: Definition, Graph, and Examples Definition of Negative Slope

A negative slope refers to the slope of a line that trends downwards when moving from left to right on a graph. In mathematical terms, the slope of a line is the change in y-coordinate with respect to the change in x-coordinate, expressed as ΔyΔx\frac{\Delta y}{\Delta x}ΔxΔy​ or RiseRun\frac{Rise}{Run}RunRise​. A negative slope indicates that two variables are negatively related - when x increases, y decreases. A line with a negative slope makes an obtuse angle (greater than 90 degrees) with the positive x-axis in the counterclockwise direction.

There are four types of slopes in mathematics: positive, negative, zero, and undefined. A positive slope rises up as we move from left to right, making an acute angle with the positive x-axis. A negative slope sinks down when moving left to right. Zero slope occurs when a line makes a 0-degree angle with the positive x-axis, creating a horizontal line parallel to the x-axis. An undefined slope happens when a line makes a 90-degree angle with the positive x-axis, creating a vertical line parallel to the y-axis.

Examples of Negative Slope Example 1: Finding if a Line Through Two Points Has a Negative Slope Problem:

Find whether the line passing through the points (5,2)(5,2)(5,2) and (2,−5)(2,-5)(2,−5) has a negative slope.

Step-by-step solution:

Step 1, Recall the slope formula. The slope can be found using the formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}m=x2​−x1​y2​−y1​​.

Step 2, Assign the points. Let's call (x1,y1)=(5,2)(x_1, y_1) = (5, 2)(x1​,y1​)=(5,2) and (x2,y2)=(2,−5)(x_2, y_2) = (2, -5)(x2​,y2​)=(2,−5).

Step 3, Find the difference in y-coordinates. Calculate y2−y1=(−5)−2=−7y_2 - y_1 = (-5) - 2 = -7y2​−y1​=(−5)−2=−7.

Step 4, Find the difference in x-coordinates. Calculate x2−x1=2−5=−3x_2 - x_1 = 2 - 5 = -3x2​−x1​=2−5=−3.

Step 5, Calculate the slope using the formula. m=−7−3=73m = \frac{-7}{-3} = \frac{7}{3}m=−3−7​=37​, which is a positive value.

Step 6, Make your conclusion. Since the slope is positive (not negative), the line passing through points (5,2)(5,2)(5,2) and (2,−5)(2,-5)(2,−5) does not have a negative slope.

Example 2: Determining if an Equation Represents a Line with Negative Slope Problem:

Show that the line with equation 5x+2y=55x + 2y = 55x+2y=5 has a negative slope.

Step-by-step solution:

Step 1, Start with the given equation of the line: 5x+2y=55x + 2y = 55x+2y=5.

Step 2, Rearrange to isolate yyy. First, subtract 5x5x5x from both sides: 2y=−5x+52y = -5x + 52y=−5x+5.

Step 3, Divide both sides by 222 to solve for yyy: y=−52x+52y = \frac{-5}{2}x + \frac{5}{2}y=2−5​x+25​.

Step 4, Compare with the slope-intercept form. The standard form is y=mx+cy = mx + cy=mx+c, where mmm is the slope. In our equation, m=−52m = \frac{-5}{2}m=2−5​.

Step 5, Draw your conclusion. Since −52\frac{-5}{2}2−5​ is negative, the line 5x+2y=55x + 2y = 55x+2y=5 has a negative slope.

Example 3: Using Angles to Determine Slope Problem:

Show that the line that makes an angle of 150150150 degrees with the positive direction of the x-axis has a negative slope.

Step-by-step solution:

Step 1, Remember the formula relating slope to angle. The slope (mmm) of a line can be calculated using m=tan⁡θm = \tan \thetam=tanθ, where θ\thetaθ is the angle the line makes with the positive x-axis.

Step 2, Identify the given angle. We're told the line makes an angle of 150150150 degrees with the positive direction of the x-axis.

Step 3, Apply the formula with the given angle. m=tan⁡150°m = \tan 150°m=tan150°.

Step 4, Break down the tangent calculation. We can rewrite tan⁡150°\tan 150°tan150° as tan⁡(90°+60°)\tan(90° + 60°)tan(90°+60°).

Step 5, Use the tangent formula for angles greater than 90°90°90°. This gives us tan⁡150°=−tan⁡30°=−13=−33\tan 150° = -\tan 30° = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}tan150°=−tan30°=−3​1​=−33​​.

Step 6, State your conclusion. Since the slope m=−33m = -\frac{\sqrt{3}}{3}m=−33​​ is negative, the line with an angle of 150150150 degrees along the positive direction of the x-axis indeed has a negative slope.

Comments(1)TTherapistVinceNovember 4, 2025This glossary page on negative slope is great! I've used it to help my students understand, and they finally get the concept. Thanks!

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