Question Video: Finding the Equations of Two Straight Lines in Two Dimensions Using the Tangent of the Angle between Them | Nagwa Question Video: Finding the Equations of Two Straight Lines in Two Dimensions Using the Tangent of the Angle between Them | Nagwa

Question Video: Finding the Equations of Two Straight Lines in Two Dimensions Using the Tangent of the Angle between Them Mathematics • First Year of Secondary School

Let πœƒ be the angle between two lines that pass through (4, βˆ’2). If tan πœƒ = 1/21 and the slope of the lines are π‘š and 4/5 π‘š, with π‘š > 0, find the equations of these lines

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Video Transcript

Let πœƒ be the angle between two lines that pass through four, negative two. If tan of πœƒ equals one over 21 and the slope of the lines are π‘š and four-fifths π‘š, with π‘š greater than zero, find the equations of these lines.

Let’s recall that by the angle between two lines we mean the smaller of the two angles. We can define the two lines as 𝐿 sub one and 𝐿 sub two. We are given some information about these two lines, firstly that they pass through this point four, negative two. And we’re also told that the tangent of the angle between them is equal to one over 21. We are further given that the slopes, or gradients, of these two lines are π‘š and four-fifths π‘š, where π‘š is greater than zero.

Before we can find the equations of 𝐿 sub one and 𝐿 sub two, the first thing we can do is work out the value of π‘š. To do this, we can recall that the formula for the tan of the acute angle πœƒ between two lines is given as tan of πœƒ equals the absolute value of π‘š sub one minus π‘š sub two over one plus π‘š sub one π‘š sub two, where π‘š sub one and π‘š sub two are the slopes of the two lines. We can therefore substitute in the given values for the tan of πœƒ and π‘š sub one and π‘š sub two. This gives us one over 21 is equal to the absolute value of π‘š minus four-fifths π‘š over one plus π‘š times four-fifths π‘š.

We can then simplify the right-hand side and multiply the numerator and denominator by five to give us that one over 21 is equal to π‘š over five plus four π‘š squared. And we can note that since π‘š is positive, then the right-hand side of this equation is also positive. By taking the cross product, we have five plus four π‘š squared is equal to 21π‘š. Then, by further rearranging, we can see that we have a quadratic equation in π‘š that we can solve for π‘š. At this point, we might realize that we can’t factorize this equation, so we can use a method, for example, the quadratic formula, to help us do this.

The quadratic formula states that π‘₯ is equal to negative 𝑏 plus or minus the square root of 𝑏 squared minus four π‘Žπ‘ all over two π‘Ž for equations of the form π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐 is equal to zero. In our equation, we have the variable π‘š instead of π‘₯. But we can see from the equation that π‘Ž is equal to four, 𝑏 is equal to negative 21, and 𝑐 is equal to five. So we can substitute these values into the quadratic formula and simplify. When we do this, we produce two different values of π‘š. And when we’re solving this, it’s helpful to remember that the square root of 361 is 19. Therefore, the two values of π‘š are five and one-quarter.

It’s important to realize that at this point even though we have found two different values of the slope π‘š, this isn’t the same as the slopes of the original lines, which we are told were π‘š and four-fifths π‘š. In fact, what we will end up with are two different pairs of lines that meet all the criteria in the question. We can now clear some space to find the two different pairs of equations for 𝐿 sub one and 𝐿 sub two.

Given that we have a point that these lines pass through and the slope of the line, we can use the point–slope form of the equation. 𝑦 minus 𝑦 sub zero equals π‘š times π‘₯ minus π‘₯ sub zero, where π‘š is the slope and the line passes through π‘₯ sub zero, 𝑦 sub zero. Notice that we will have to be a little bit careful when we’re using this point–slope form that we don’t get confused with this general letter π‘š in the formula and the variable π‘š and four-fifths π‘š that we were using in the question. Let’s take the instance of π‘š which is just equal to five. So we know that one of our lines has a slope which is just equal to π‘š, which is five.

Given that this line passes through the point four, negative two, when we use the point–slope form, we’ll have 𝑦 minus negative two is equal to five times π‘₯ minus four. Simplifying this and then collecting the like terms, we have that the first line is negative five π‘₯ plus 𝑦 plus 22 is equal to zero. We still need to find the second line when π‘š is equal to five. But this time, we know that the slope is actually equal to four-fifths of π‘š. Since π‘š is equal to five, then the slope of the second line will be equal to four. It’s still going through the point four, negative two. So this time on the right-hand side, we’ll be multiplying outside these parentheses by the value four. Simplifying, we have the equation of the line as four π‘₯ minus 𝑦 minus 18 is equal to zero.

These two lines form one possible set of lines, which meet all the given criteria in the question. However, we also know that these criteria will still be met when the value of π‘š is equal to one-quarter. So let’s use the point–slope form of the line again to work out the other pair of equations. For the equation of the first line, we know that the slope is simply equal to π‘š, which is one-quarter. We can then rearrange this to give the first equation of the line, noticing that if we want to leave it in the standard form when we’re rearranging, we’ll need to multiply by four throughout the equation. We can then do the same to find the second equation.

Remember that as we’ve seen previously, in this line, the value of π‘š that we’re substituting into the point–slope form isn’t simply one-quarter because we know that the slope of the second line is four-fifths of π‘š. Four-fifths of one-quarter is one-fifth. When we substitute this value in along with the coordinate four, negative two, we get the equation of the second line. Therefore, we can give the answer of the two possible pairs of equations. π‘₯ minus four 𝑦 minus 12 equals zero and π‘₯ minus five 𝑦 minus 14 equals zero or negative five π‘₯ plus 𝑦 plus 22 equals zero and four π‘₯ minus 𝑦 minus 18 equals zero.

In order to demonstrate the difference between these two equation pairs, we could draw a quick sketch of each situation. Drawing the first pair of lines would result in this figure, and the second pair of lines would result in this figure. In both sets of graphs, the gradient, or slope, is greater than zero as it was supposed to be. And it’s also worth noting that the inverse tan of one over 21 actually produces a very small angle in degrees. It’s approximately 2.7 degrees.

So by applying the formula to find the acute angle between two straight lines and given the relationship between their slopes, we have found two different pairs of straight lines passing through the coordinate four, negative two.

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