Question Video: Using the Angle between Two Straight Lines to Solve a Problem | Nagwa Question Video: Using the Angle between Two Straight Lines to Solve a Problem | Nagwa

Question Video: Using the Angle between Two Straight Lines to Solve a Problem Mathematics • First Year of Secondary School

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If the acute angle between the straight lines whose equations are π‘˜π‘¦ βˆ’ 2π‘₯ + 19 = 0 and 9π‘₯ βˆ’ 7𝑦 βˆ’ 8 = 0 is πœ‹/4, find all the possible values of π‘˜.

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Video Transcript

If the acute angle between the straight lines whose equations are π‘˜π‘¦ minus two π‘₯ plus 19 is equal to zero and nine π‘₯ minus seven 𝑦 minus eight is equal to zero is πœ‹ by four, find all the possible values of π‘˜.

In this question, we’re given the equations of two straight lines, both given in general form. And we can see one of these straight lines has an unknown value of π‘˜. We need to determine this value of π‘˜. And to do this, we’re told that the acute angle between the two straight lines is πœ‹ by four. And it’s worth noting we need to determine all of the possible values of π‘˜, so there may be multiple correct solutions or no solutions.

To do this, let’s start by recalling how we find the acute angle between two given straight lines. We know if 𝛼 is the acute angle between two straight lines with slopes π‘š sub one and π‘š sub two, then the tan of 𝛼 will be equal to the absolute value of π‘š sub one minus π‘š sub two divided by one plus π‘š sub one times π‘š sub two. If we know the slopes of the two lines π‘š sub one and π‘š sub two, then we can solve for 𝛼 by taking the inverse tangent of both sides of the equation. However, in this case, we’re not trying to find the value of 𝛼. We’re already told that 𝛼 is πœ‹ by four. Instead, we need to find the value of π‘˜. Since we know the value of 𝛼, we could substitute this into our equation. We can then use the two given equations to find expressions for the slopes.

Let’s start with the slope of the first line. We want to find the slope of the line π‘˜π‘¦ minus two π‘₯ plus 19 is equal to zero. And we can note this line is given in general form, and it’s difficult to determine the slope of a line given in general form. So instead, let’s rearrange this into slope–intercept form. We start by adding two π‘₯ to both sides of the equation and subtracting 19 from both sides of the equation. We get π‘˜π‘¦ is equal to two π‘₯ minus 19.

Now, we’re going to divide through by π‘˜. This gives us that 𝑦 is equal to two π‘₯ over π‘˜ minus 19 over π‘˜. And it’s worth noting we’re making an assumption here. We’re assuming our value of π‘˜ is nonzero. If our value of π‘˜ is equal to zero, then this equation simplifies to give us negative two π‘₯ plus 19 is equal to zero. We can rearrange this to get π‘₯ is equal to 19 over two. In other words, this line would be a vertical line at 19 over two.

Since vertical lines do not have a defined slope, we can’t use our formula to determine the angle between two straight lines if one of the lines is vertical. So we’re going to need to treat this case separately. We can, however, determine the slope of the second line in the same way. We start by adding seven 𝑦 to both sides of the equation. We get seven 𝑦 is equal to nine π‘₯ minus eight. We then divide the equation through by seven. 𝑦 is equal to nine over seven π‘₯ minus eight over seven.

Before we use our formula, let’s now consider the case where π‘˜ is equal to zero. We’ll start by sketching our first line. It’s a vertical line at 19 over two. We know the angle between the two straight lines is πœ‹ by four; we’re told this in the question. And we can notice that this vertical line is parallel to the 𝑦-axis. So the angle between this line and any other line will be the same as the angle between that line and the 𝑦-axis since it’s a transversal of parallel lines. And in particular, we’re told that this angle is πœ‹ by four. So we can also see that the angle it makes with the positive π‘₯-axis will also be πœ‹ by four. And the only line which makes an angle of πœ‹ by four in this manner with the positive π‘₯-axis will have slope one.

It is worth noting here we’ve made one small assumption. We’ve assumed that our line is positive slope. We get a very similar story if we assumed our line had negative slope. Instead, we would have found that the angle with the negative π‘₯-axis was πœ‹ by four. So it would have had slope negative one. And in either case, we can show this is not what’s happening in this case. Since we’ve already found the slope of the second line, the slope of the second line in slope–intercept form is its coefficient of π‘₯, nine over seven. And this is not negative one or positive one. So the value of π‘˜ cannot be equal to zero. So we’ve shown that π‘˜ is nonzero, so let’s clear some space and carry on with this method.

We can find the slope of the two lines as the coefficients of π‘₯. π‘š sub one is two over π‘˜ and π‘š sub two is nine over seven. We can now substitute these into our formula along with the fact that 𝛼 is equal to πœ‹ by four. Substituting these in, we get the tan of πœ‹ by four is equal to the absolute value of two over π‘˜ minus nine over seven divided by one plus two over π‘˜ times nine over seven. This is now an equation entirely in terms of π‘˜. So we can try and solve this for π‘˜.

So let’s start by simplifying the right-hand side of this equation. Let’s start with the numerator. We want to multiply two over π‘˜ by seven over seven and nine over seven by π‘˜ divided by π‘˜. Doing this and simplifying, we get 14 minus nine π‘˜ all divided by seven π‘˜. We can also simplify the denominator. Two over π‘˜ multiplied by nine over seven is equal to 18 divided by seven π‘˜. Therefore, the right-hand side of this equation simplifies to give us the absolute value of 14 minus nine π‘˜ over seven π‘˜ divided by one plus 18 over seven π‘˜. We can evaluate the left-hand side of the equation. The tan of πœ‹ by four is just equal to one.

We still need to simplify the right-hand side of this equation. We’ll do this by multiplying both the numerator and denominator by seven π‘˜. Doing this and simplifying, we get that one is equal to the absolute value of 14 minus nine π‘˜ divided by seven π‘˜ plus 18. We can now solve this as an absolute value equation. We recall if the absolute value of π‘₯ is equal to one, then either π‘₯ is equal to one or π‘₯ is equal to negative one. This gives us two equations. Either 14 minus nine π‘˜ over seven π‘˜ plus 18 is equal to one or 14 minus nine π‘˜ over seven π‘˜ plus 18 is equal to negative one.

We can solve both of these equations for π‘˜. In the first equation, we multiply through by seven π‘˜ plus 18. This gives us seven π‘˜ plus 18 is equal to 14 minus nine π‘˜. Now, we can add nine π‘˜ to both sides of the equation and subtract 18 from both sides of the equation. This gives us that 16π‘˜ is equal to negative four. Finally, we’ll divide the equation through by 16. This gives us that π‘˜ is equal to negative four over 16, which simplifies to negative one-quarter. We can do the same in the second case. We multiply through by seven π‘˜ plus 18. This gives us that negative seven π‘˜ minus 18 is equal to 14 minus nine π‘˜.

We can now solve this equation by adding nine π‘˜ to both sides of the equation and adding 18 to both sides of the equation. This gives us that two π‘˜ is equal to 32. And we can solve for π‘˜ by dividing both sides of the equation through by two. We get that π‘˜ is equal to 16, and this gives us our final answer. If the acute angle between the two given straight lines is πœ‹ by four, then there are only two possible values of π‘˜. Either π‘˜ is negative one-quarter or π‘˜ is equal to 16.

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