A line passing through eight, two makes an angle 𝜃 with the line six 𝑥 plus four 𝑦 plus nine equals zero, and tan of 𝜃 equals 15 over 13. What is the equation of this line?
Let’s think about what we know. We know that line one passes through the point eight, two. We know that the equation for our second line is six 𝑥 plus four 𝑦 plus nine equals zero. We know that these two lines intersect at the angle 𝜃. And we should remember something about the tangent of this angle. It’s equal to the absolute value of the slope of line one minus the slope of line two over one plus the slope of line one times the slope of line two. And how many of these values do we already know? We know that tan of 𝜃 equals 15 over 13. We don’t know the slope of the second line, but we can find it using the information we were given.
This means we will be able to solve to find 𝑚 one, the slope of the first line. If we find the slope 𝑚 one, because we were already given a point, we’ll have the slope of line one and a point that line one passes through, which will allow us to find the equation of this line, line one. So let’s get started. The tan of 𝜃 equals the absolute value of 𝑚 one minus 𝑚 two over one plus 𝑚 one times 𝑚 two. The tan of 𝜃 equals 15 over 13. Before we can plug in any other information, we need to solve for the slope of line two.
To isolate the slope easily from this equation, we would prefer to rewrite it in slope intercept form, which is 𝑦 equals 𝑚𝑥 plus 𝑏. And in this form, 𝑚 is the slope. To find this form, we need to get 𝑦 by itself. And that means we need to subtract six 𝑥 and nine from both sides of the equation. When we do that, we’ll be left with four 𝑦 equals negative six 𝑥 minus nine. But we need 𝑦 completely by itself. Its coefficient needs to be one. And so we divide everything by four. Four 𝑦 over four equals 𝑦. Negative six over four 𝑥 can be simplified to negative three-halves 𝑥 minus nine-fourths. The slope of line two is then negative three-halves. And so we’ll say 𝑚 two equals negative three-halves.
And so we plug in negative three-halves into our equation everywhere we see 𝑚 two. At this point, we can simplify our numerator. We’re taking 𝑚 one and we’re subtracting negative three-halves. We simplify that to say 𝑚 one plus three-halves. And then our denominator will be one minus three-halves times 𝑚 one. Because we’re dealing with absolute value, we will be dealing with two different solutions, a positive solution and a negative solution. The positive solution, 15 over 13, equals 𝑚 one plus three-halves over one minus three-halves 𝑚 one. And the negative solution, negative 15 over 13, equals 𝑚 one plus three-halves over one minus three-halves 𝑚 one.
Let’s focus on the positive solution first. We first need to cross multiply the numerators and the denominators, which will be the numerator 15 multiplied by the denominator one minus three-halves 𝑚 one is equal to the denominator 13 multiplied by the numerator 𝑚 one plus three-halves. We distribute the multiplication across the brackets. 15 times one is 15. 15 times negative three-halves 𝑚 one is negative forty-five halves 𝑚 one, which is equal to 13 times 𝑚 one, 13 𝑚 one, plus 13 times three-halves, 39 over two. To solve for 𝑚 one, we need to get 𝑚 one on the same side of the equation. And so we subtract 13 𝑚 one from both sides of the equation. Negative 45 halves 𝑚 one minus 13 𝑚 one will equal negative seventy-one halves 𝑚 one.
We also want to move this positive 15 to the other side of the equation. And so we subtract 15 from both sides. On the left, 15 minus 15 equals zero and thirty-nine halves minus 15 equals nine-halves. Since we have a denominator of two on both sides of the equation, we can multiply both sides by two, which will leave us with negative 71 𝑚 one equals nine. To find 𝑚 one, we divide both sides of the equation by negative 71. And we see that 𝑚 one equals negative nine over 71. This is one possible option for the slope of line one. But because we had an equation with absolute value, we need to also consider the negative solution.
To deal with negative 15 13ths, we can say negative 15 over 13, and we’ll follow the same cross-multiplication procedure. Multiply the numerator, negative 15, by the denominator, one minus three-halves 𝑚 one. And then we multiplied the denominator, 13, by the numerator, 𝑚 one plus three-halves. First, we distribute the negative 15 across the brackets, which gives us negative 15. Negative 15 times negative three-halves 𝑚 one is positive forty-five halves 𝑚 one. We’ll distribute the 13 over the 𝑚 one and the three-halves, which is the same as the first time, 13𝑚 one plus thirty-nine halves.
Again, to solve for 𝑚 one, we’ll need to get them both on the same side of the equation. And so we subtract 13𝑚 one from both sides. Be careful here; this time, we’re subtracting 13𝑚 one from positive 45 over two 𝑚 one, instead of negative 45 over two 𝑚 one. Forty-five halves minus 13 equals nineteen-halves. We have nineteen-halves 𝑚 one. This time we have a negative 15 on the left. And so we add 15 to both sides. Negative 15 plus 15 equals zero. Thirty-nine halves plus 15 equals sixty-nine halves.
Again, we’ll multiply the entire equation by two, which leaves us with 19𝑚 one equals 69. And we divide both sides of the equation by 19 to see that 𝑚 one is also equal to 69 over 19. So far, we can say that line one has a slope of either negative nine over 71 or 69 over 19. But we’re looking for an equation for this line. Since we have the slope of line one and a point for line one, we can use the point-slope formula to find the equations for this line. This formula says 𝑦 minus 𝑦 one equals 𝑚 times 𝑥 minus 𝑥 one where 𝑥 one, 𝑦 one is your point. And of course, 𝑚 is your slope. Our point is eight, two. And we have two different slopes, which means we’ll have to use this formula twice, the first time for 𝑚 equals negative nine over 71 and the second time for 69 over 19.
With our point in our slope, we can say 𝑦 minus two equals negative nine over 71 times 𝑥 minus eight. We distribute our negative nine over 71. And we get 𝑦 minus two is equal to negative nine over 71𝑥 plus 72 over 71. From there, we add two to both sides. And we’ll get 𝑦 equals negative nine over 71𝑥 plus 214 over 71. This is one equation for line one written in slope intercept form. Sometimes it’s useful to not have a fraction in the denominator. And so we could rearrange this equation. By multiplying everything by 71, we would then have 71𝑦 equals negative nine 𝑥 plus 214. The equation for line two was given to us as an equation set equal to zero. And so we could also subtract 71𝑦 from both sides. And we could have the equation zero equals negative nine 𝑥 minus 71𝑦 plus 214.
In this case, we have negative values as our coefficients for 𝑥 and 𝑦. If we didn’t want that, we could rearrange it by multiplying the equation through by negative one to give us positive nine 𝑥 plus 71𝑦 minus 214. All four of these formats are valid ways of expressing the equation of this line. You don’t need all four, but it’s helpful to know how to rearrange them to find equivalent expressions of the same line. We now do this process one more time using the slope of 69 over 19. The point-slope formula gives us 𝑦 minus two equals 69 over 19 times 𝑥 minus eight. We distribute the 69 over 19 which gives us 𝑦 minus two equals 69 over 19𝑥 minus 552 over 19. And then we add two to both sides. And then we have 𝑦 equals 69 over 19𝑥 minus 514 over 19.
This is the slope-intercept form of the second option for line one. If we multiply that whole equation by 19, we would have the equivalent equation: 19𝑦 equals 69𝑥 minus 514. And if we subtracted 19𝑦 from both sides, we would have the equivalent equation zero equals 69𝑥 minus 19𝑦 minus 514. If we multiplied that equation by negative one, we would get zero equals negative 69𝑥 plus 19𝑦 plus 514. What all this information tells us is that there are two lines that pass through the point eight, two and make the angle tan of 𝜃 equals 15 over 13 with the line six 𝑥 plus four 𝑦 plus nine equals zero. So you just need to select one equation from each column and you’ll have an equation for the two lines that fit this criteria.