Lesson Video: Slopes of Parallel and Perpendicular Lines | Nagwa Lesson Video: Slopes of Parallel and Perpendicular Lines | Nagwa

Lesson Video: Slopes of Parallel and Perpendicular Lines Mathematics • Third Year of Preparatory School

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In this video, we will learn how to use the concept of slopes to determine whether two lines are parallel or perpendicular and use these geometric relationships to solve problems.

16:36

Video Transcript

In this video, we will learn how to use the concept of slope to determine whether two lines are parallel or perpendicular. And then, we’ll see how we can use these geometric relationships to solve problems.

The slope of a line is a very important feature of a line, and it describes how steep a line is. The slope of a line can be calculated from any two distinct points on a line. In general, we can say that if there are two points on a straight line with the coordinates π‘₯ sub zero, 𝑦 sub zero and π‘₯ sub one, 𝑦 sub one, then we can calculate the slope, often referred to using the letter π‘š, as 𝑦 sub one minus 𝑦 sub zero over π‘₯ sub one minus π‘₯ sub zero. To find the slope, we’re really dividing the vertical displacement, that’s the change in 𝑦, by the horizontal displacement, or the change in π‘₯. As a recap of how this works in practice, let’s say that we have the two coordinates four, six and 12, 10. We can define four, six to have the π‘₯ sub zero, 𝑦 sub zero values, although it wouldn’t matter if we define these with the π‘₯ sub one, 𝑦 sub one values.

Substituting these into the slope formula, we would have that π‘š is equal to 10 minus six over 12 minus four. This would simplify to four over eight, which in turn simplifies to one-half. The slope of this line is one-half. We often think of this calculation in very much algebraic terms, but let’s take a closer look at the geometry involved. When we’re finding the slope of a line, we are creating a right triangle. The lengths of the two shorter sides are the horizontal and vertical displacements. So when we think about slope in terms of right triangles, then we can use results that we know from trigonometry to understand other properties of the straight line.

One property that we’re often interested in is the acute angle that the line makes with the horizontal axis, which we can label as 𝛼. The important thing to note here is that the angle 𝛼 made between line 𝐴𝐡 and this horizontal line will be the same as the angle 𝛼 made between the line 𝐴𝐡 and the horizontal axis because these two horizontal lines are parallel. So we can find the angle between the straight line and the horizontal axis by finding the angle 𝛼. And as already mentioned, we can do this by using trigonometry. In this problem, we have the angle 𝛼, we have the side opposite the angle, and we have the side adjacent to the angle.

We can therefore use the fact that the tangent is the ratio of the opposite side and the adjacent side in a right triangle. And so we have that tan of 𝛼 is equal to 𝑦 sub one minus 𝑦 sub zero over π‘₯ sub one minus π‘₯ sub zero. In other words, tan of the angle 𝛼 is simply equal to π‘š, where π‘š is the slope of the straight line. So remember that we worked out the slope of this example line as one-half. If we then wanted to work out the angle 𝛼, we know that tan of 𝛼 is equal to one-half. Therefore, 𝛼 is equal to arctan of one-half. In degrees, this is approximately 26.57 degrees to two decimal places. We can also use this approach to find the angle between a straight line and the horizontal axis when the angle is not acute.

We know that it is possible for the slope of a straight line to be negative, which happens when 𝑦 sub one minus 𝑦 sub zero and π‘₯ sub one minus π‘₯ sub zero are of an opposite sign. In this case, the straight line will go downwards from left to right. And then the positive angle, which is the angle measured clockwise, between the positive direction of the π‘₯-axis and the straight line is then obtuse. As we can observe, by using a calculator, the tangent of an obtuse angle is negative. And so the relationship we have found between the slope of a line and the tangent of the positive angle the line makes with the positive direction of the π‘₯-axis also holds for obtuse angles.

We can now make a more formal note of what we have learned. Firstly, we know that the slope π‘š between two coordinates π‘₯ sub zero, 𝑦 sub zero and π‘₯ sub one, 𝑦 sub one is given as π‘š equals 𝑦 sub one minus 𝑦 sub zero over π‘₯ sub one minus π‘₯ sub zero. Furthermore, the slope is equal to the tangent of the positive angle made between the straight line and the positive direction of the π‘₯-axis such that π‘š is equal to tan of 𝛼. The angle 𝛼 is measured from the positive π‘₯-axis to the line in a counterclockwise direction. An acute angle has a positive tangent, whereas an obtuse angle has a negative tangent. And a final note that because the tan of an angle of 90 degrees is undefined, then vertical lines are said to have an undefined slope.

We’ll now take a look at an example where we find the slope of a line given the angle it makes with the horizontal axis.

Find, to the nearest two decimal places, the slope of the line that makes a positive angle of 60 degrees with the positive direction of the π‘₯ axis.

We can begin this problem by visualizing a line that makes an angle of 60 degrees with the positive direction of the π‘₯-axis. In order to answer this problem, we will also need to remember that the slope of a straight line π‘š is equal to the tangent of the positive angle made between the straight line and the positive direction of the π‘₯-axis. In this question, that angle would be 60 degrees. So we would have that π‘š is equal to tan of 60 degrees. tan of 60 is equal to root three, but as a decimal it would be 1.732 and so on. Rounded to the nearest two decimal places then, we can say that the slope of the line is 1.73.

We will now move on to looking at parallel and perpendicular lines. Let’s consider the fact that two lines meet at a point unless they are parallel or coincident. Coincident lines will lie exactly on top of one another. We know that two lines are parallel if they have the same slope. And from what we have just seen in this video, we can now add that lines are parallel if they make the same angle with the positive direction of the π‘₯-axis. If lines have a slope of zero, then they are parallel to the π‘₯-axis and parallel to each other even if they don’t cross the π‘₯-axis. Two lines are parallel but not coincident when they have the same slope but not the same 𝑦-intercept, as shown in this diagram.

We should recall that perpendicular lines meet at a point and make an angle of 90 degrees with each other. We will now see what this means for the slopes of two perpendicular lines. Let’s take these two lines, which have slopes of π‘š sub one and π‘š sub two. They make angles of 𝛼 and 𝛽, respectively, with the positive direction of the π‘₯-axis.

One of the things we can say is that because the lines are perpendicular, then 𝛽 is equal to 𝛼 plus 90 degrees. A property of the tangent function is that tan of 𝛼 is equal to negative one over tan 𝛼 plus 90 degrees. Then, combining these two equations, we have that tan of 𝛼 is equal to negative one over tan of 𝛽. And then, because we know that π‘š sub one is equal to tan of 𝛼 and π‘š sub two is equal to tan of 𝛽, we have that π‘š sub one is equal to negative one over π‘š sub two. Alternatively, this can be written as π‘š sub one, π‘š sub two is equal to negative one.

You might wonder why this is important, but what we have really demonstrated here is that the product of the perpendicular slopes is negative one. This is a very important property of perpendicular lines, but note that if a line is horizontal, the slope is zero. For example, if π‘š sub two is zero, then to find π‘š sub one, we would be attempting to divide by zero. This would give us an undefined value, but of course the slope of a vertical line is undefined. This makes sense because we know that a vertical line is perpendicular to a horizontal line. But we can’t automatically use this algebraic fact that π‘š sub one times π‘š sub two equals negative one with horizontal and vertical lines.

We can now make a quick summary of the conditions for parallel and perpendicular lines. We can identify parallel lines as having the same slope and a different 𝑦- intercept. Then, lines which are identical have the same slope and the same 𝑦-intercept. And then, when the product of the slopes is equal to negative one, then the two lines are perpendicular. And as previously noted, if the slope of the line is zero, then the straight line is horizontal. Any line which is perpendicular to this would not have a defined slope.

In the next example, we’ll see how we can find the slope of a straight line given the slope of a perpendicular one.

If line 𝐴𝐡 is perpendicular to line 𝐢𝐷 and the slope of line 𝐴𝐡 equals two-fifths, find the slope of line 𝐢𝐷.

Here, we are told that we have two perpendicular lines 𝐴𝐡 and 𝐢𝐷. Knowing that two lines are perpendicular means that we know something about the relationship between their slopes. If we define line 𝐴𝐡 to have a slope of π‘š sub one and line 𝐢𝐷 to have a slope of π‘š sub two, then we know that π‘š sub two is equal to negative one over π‘š sub one. Given that the slope π‘š sub one of line 𝐴𝐡 is two-fifths, then π‘š sub two is equal to negative one over two-fifths. This simplifies to negative five over two. Because these two lines are perpendicular, their slopes will be the negative reciprocal of one another. And so the slope of line 𝐢𝐷 is negative five over two.

In the next example, we’ll identify the relationship between two straight lines.

Let 𝐿 be the line through the points negative seven, negative seven and negative nine, six and 𝑀 the line through one, one and 14, three. Which of the following is true about the lines 𝐿 and 𝑀? Option (A) they are parallel, option (B) they are perpendicular, or option (C) they are intersecting but not perpendicular.

It might be worthwhile beginning this question with a quick sketch of the two lines through the two sets of points. When we do this, we can observe that the two lines do in fact intersect. We can therefore say that these two lines are not parallel, so we can eliminate option (A). Now, we can remember that two lines are perpendicular if they intersect or meet at right angles. From the diagram, it does appear that the two lines are at right angles. But it could be the case that the two lines are nearly perpendicular and it’s not possible to distinguish this from the diagram. Generally, it’s not a very good idea to just use a sketch to determine if lines are parallel or perpendicular. In fact, we should perform some sort of calculation.

We can recall that if two straight lines have slopes of π‘š sub one and π‘š sub two, then they are perpendicular if π‘š sub two is equal to negative one over π‘š sub one. We’ll first need to calculate the slopes of each of the lines 𝐿 and 𝑀. The slope of the line passing through two points with coordinates π‘₯ sub zero, 𝑦 sub zero and π‘₯ sub one, 𝑦 sub one is calculated as the slope π‘š is equal to 𝑦 sub one minus 𝑦 sub zero over π‘₯ sub one minus π‘₯ sub zero. For line 𝐿 then, its slope π‘š sub one is equal to six minus negative seven over negative nine minus negative seven, which simplifies to negative 13 over two.

Now, let’s find the slope of the line 𝑀. Its slope π‘š sub two will be calculated as three minus one over 14 minus one, and this is equal to two over 13. Now, we can check if π‘š sub two is equal to negative one over π‘š sub one. If we didn’t know the value of π‘š sub two, we could find a perpendicular line to the line 𝐿 by taking π‘š sub two and setting it equal to negative one over negative 13 over two. And this would indeed give us a value of two thirteenths for π‘š sub two. We can therefore give the answer that the statement which is true about the lines 𝐿 and 𝑀 is option (B). They are perpendicular.

We will now summarize the key points of this video. The slope π‘š of a straight line passing through π‘₯ sub zero, 𝑦 sub zero and π‘₯ sub one, 𝑦 sub one is given by π‘š equals 𝑦 sub one minus 𝑦 sub zero over π‘₯ sub one minus π‘₯ sub zero. The angle 𝛼 is measured from the horizontal axis, rotating counterclockwise until meeting the straight line. For π‘š being the slope of a straight line, the following results hold. For π‘š is greater than or equal to zero, the angle 𝛼 between this straight line and the horizontal axis is expressed as 𝛼 equals arctan of π‘š. For π‘š is less than zero, the angle 𝛼 between this straight line and the horizontal axis is expressed as 180 degrees plus arctan of π‘š. And for vertical lines, 𝛼 equals 90 degrees.

We also saw that if we take two straight lines with slopes π‘š sub one and π‘š sub two and a 𝑦-intercept 𝑐 one and 𝑐 two, then if π‘š sub one equals π‘š sub two and 𝑐 sub one is not equal to 𝑐 sub two, then the two lines are distinct and parallel. This means that the lines never meet and that they make the same angle with the horizontal axis. But if π‘š sub one equals π‘š sub two and 𝑐 sub one equals 𝑐 sub two, then the lines are coincident or identical. But if π‘š sub one times π‘š sub two equals negative one, then the two lines are perpendicular.

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