Video: Finding the Equation of a Straight Line

Find the equation of the line perpendicular to βˆ’6π‘₯ βˆ’ 𝑦 + 8 = 0 and passing through the intersection of the lines βˆ’4π‘₯ βˆ’ 𝑦 βˆ’ 3 = 0 and βˆ’3π‘₯ + 8𝑦 βˆ’ 1 = 0.

06:32

Video Transcript

Find the equation of the line perpendicular to negative six π‘₯ minus 𝑦 plus eight is equal to zero and passing through the intersection of the lines negative four π‘₯ minus 𝑦 minus three equals zero and negative three π‘₯ plus eight 𝑦 minus one is equal to zero.

Within this question, we’re given two key pieces of information about the line whose equation we want to find. Firstly, we’re told something about its slope. It’s perpendicular to the line negative six π‘₯ minus 𝑦 plus eight is equal to zero. Secondly, we’re told about a point that lies on this line. It passes through the intersection of the lines whose equations are given.

To answer this question, we’ll need to find both the slope of the line π‘š and the coordinates of this point that lies on the line π‘₯ one, 𝑦 one so that then we can apply the point-slope method to find the equation of this line 𝑦 minus 𝑦 one is equal to π‘š π‘₯ minus π‘₯ one.

Let’s concentrate on finding π‘š β€” the slope of the line first. We know that if two lines are perpendicular to each other, then the product of their slopes, which are referred to as π‘š one and π‘š two, is equal to negative one. If we take the equation of the line that’s perpendicular to our line and add 𝑦 to each side, then we have negative six π‘₯ plus eight is equal to 𝑦.

Comparing this with 𝑦 equals π‘šπ‘₯ plus 𝑐, the slope-intercept form of the equation of a straight line, we see that the slope of this line is negative six. To find the slope of the perpendicular line, we need to divide negative one by negative six. The slope of the perpendicular line β€” our line β€” is one-sixth.

So we found the slope of the line and I’ve substituted it into the beginnings of our equation for the line. I’m going to delete the working out for this part from the screen in order to make way for the next stage of working out. So if you need to jot any of it down, pause the video and do so now.

Next, we need to find the coordinates of the point of intersection of the two lines given. To do this, we need to solve their equations simultaneously. I’ve labelled the equations as equation one and equation two. I know now that equation one can be rearranged to give an expression for 𝑦 in terms of π‘₯. Adding 𝑦 to both sides of the equation gives 𝑦 is equal to negative four π‘₯ minus three.

The next step is to substitute this expression for 𝑦 into equation two. This gives negative three π‘₯ plus eight lots of negative four π‘₯ minus three minus one is equal to zero. And now, I have a linear equation in π‘₯ which I can solve. First, I expand the bracket to give negative three π‘₯ minus 32π‘₯ minus 24 minus one is equal to zero. Now, I group like terms together giving minus 35π‘₯ minus 25 is equal to zero.

The next step is to add 25 to both sides of the equation giving negative 35π‘₯ is equal to 25. Next, I divide both sides of the equation by negative 35, giving π‘₯ is equal to 25 over negative 35. Both the numerator and denominator in this fraction can be cancelled by a factor of five. And we have a positive divided by a negative, giving a negative overall. We have the π‘₯ is equal to negative five over seven.

So we found the value of π‘₯ and now we need to find the value of 𝑦. To do so, we can substitute into the rearranged form of equation one. This gives 𝑦 is equal to negative four multiplied by negative five over seven minus three. Negative four multiplied by negative five over seven is positive 20 over seven. And if I want to write negative three as a fraction with a denominator of seven, it’s equal to negative 21 over seven. This then simplifies to give negative one over seven.

So by solving the equations of the two lines simultaneously, we found the coordinates of the point of intersection and hence the coordinates of the point that lies on the line we’re looking for. We can substitute the two values as π‘₯ one and 𝑦 one in the equation of our line.

Now again, I’m going to delete some of the working out on screen in order to complete the final part of the question. So pause the video now if you need to write anything down. We found the slope of the line and the coordinates of the point that lies on the line. And therefore, we have the equation of our line 𝑦 minus negative one over seven is equal to one-sixth of π‘₯ minus negative five over seven.

We now need to rearrange the equation of our line. The two negatives on the left-hand side and the two negatives on the right-hand side can each be simplified to a plus. So we have 𝑦 plus one-seventh is equal to one-sixth of π‘₯ plus five-sevenths. Now, there’re a lot of fractions in our equation at the moment and we prefer to be working with integers. So I’m going to eliminate these fractions.

First of all, I’m going to multiply every term in the equation by six. This gives six 𝑦 plus six-seventh is equal to π‘₯ plus five-sevenths. Now, I’ve still got fractions involved. So next, I’m gonna multiply every term in the equation by seven. This gives 42𝑦 plus six is equal to seven π‘₯ plus five and I have eliminated all of the fractions.

Finally, I’m going to group all of the terms on the same side of the equation and I’m going to choose to group on the right-hand side. To do so, I need to subtract both 42𝑦 and six from both sides of the equation. This gives zero is equal to seven π‘₯ minus 42𝑦 minus one.

By swapping the zero into the right-hand side of the equation, which is a little more conventional, we have the equation of the line we were looking for: seven π‘₯ minus 42𝑦 minus one is equal to zero.

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