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In this video, we’re going to look at how to find the shortest distance from a point to a straight line in the coordinate plane.
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So this is the situation that we’re interested in.
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We have a coordinate plane and a straight line and a point.
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And we’re looking to find the shortest distance between the point and the line.
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Now the first question might be, “Well, how do I determine in what direction that distance should be?”
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There are lots of different possible distances from the point to the line, depending which point on the line you choose to connect to.
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A key fact that you need to be aware of is that the shortest distance from a point to a line is the perpendicular distance.
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So when we refer to the distance between a point and a line, it’s this distance here that we’re looking to calculate.
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So I’ve chosen to label the equation of the line in the form 𝑎𝑥 plus 𝑏𝑦 plus 𝑐 is equal to 0.
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And I’ve chosen to label the point they we’re interested in as having the coordinates 𝑥 one, 𝑦 one.
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There is a formula that we can use in order to calculate the distance between the point and the line.
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And it’s this formula here.
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𝑙, which represents the distance, is equal to the absolute value or the modulus of 𝑎𝑥 one plus 𝑏𝑦 one plus 𝑐, all divided by the square root of 𝑎 squared plus 𝑏 squared.
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Those modulus signs in the numerator mean you take the absolute value of a quantity.
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So if it’s a positive value, then it’s just that value.
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But if it’s a negative value, then you multiply it by negative one.
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So, for example, the modulus of six is six and the modulus of negative six is also six.
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Now this formula is a handy shortcut if you can remember it.
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But there is also a more complete method that you may want to use, which is perhaps a little bit more intuitive.
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This method doesn’t just consider that small distance 𝑙, but the full line which it’s part of.
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Now this line is perpendicular to the line that you would be given in the question.
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So the first step is to find its equation.
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Remember, you know the coordinates of one point on this line, 𝑥 one, 𝑦 one.
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You also know the slope of this line because it’s perpendicular to the line 𝑎𝑥 plus 𝑏𝑦 plus 𝑐 equals zero.
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You’d need to use the relationship between the slopes of perpendicular lines, that is, that they multiply to negative one.
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Once you’ve found the equation of this line, you would then solve the equation simultaneously with the equation of the first line in order to find the point of intersection of the two lines, which will give the coordinates 𝑥 two, 𝑦 two.
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Once you have the coordinates of this point, we then use the distance formula in order to calculate the distance 𝑙, the square root of 𝑥 two minus 𝑥 one all squared plus 𝑦 two minus 𝑦 one all squared.
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That’s just an application of the Pythagorean theorem.
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So as I mentioned, this method will certainly be longer and perhaps more complex than using the formula that I mentioned previously.
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But this is the method behind what that formula was doing.
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So we’ll see both methods within this video.
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Find the length of the perpendicular drawn from the point 𝐴 one, nine to the straight line negative five 𝑥 plus 12𝑦 plus 13 equals zero.
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So we’re going to answer this question using the formula for calculating the distance between a point and a straight line.
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So the formula is this.
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If I have the straight line with equation 𝑎𝑥 plus 𝑏𝑦 plus 𝑐 is equal to zero and I have a point with coordinates 𝑥 one, 𝑦 one.
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Then the perpendicular distance between them, 𝑙, is given by the modulus of 𝑎𝑥 one plus 𝑏𝑦 one plus 𝑐, all divided by the square root of 𝑎 squared plus 𝑏 squared.
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So what I need to do is determine the values of 𝑎, 𝑏, 𝑐, 𝑥 one, and 𝑦 one and then substitute them into the formula.
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Let’s look at the straight line first of all.
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I’m comparing it with 𝑎𝑥 plus 𝑏𝑦 plus 𝑐 is equal to zero.
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This shows me that 𝑎 is equal to negative five, 𝑏 is equal to 12, and 𝑐 is equal to 13.
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Now let’s look at the point 𝐴, which has coordinates one, nine.
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This tells me that 𝑥 one is equal to one and 𝑦 one is equal to nine.
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So now I have all the values I need.
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And it’s just a case of substituting them into this formula for the distance 𝑙.
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So we have that 𝑙 is equal to negative five times one plus 12 times nine plus 13, the modulus of that quantity.
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Then we’re going to divide it by the square root of negative five squared plus 12 squared.
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This gives us the modulus of negative five plus 108 plus 13 all divided by the square root of 25 plus 144.
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This gives the modulus of 116 over the square root of 169.
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Now as 116 is positive, then its modulus is just its own value.
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So the numerator will be 116.
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And in the denominator, the square root of 169 is 13 exactly.
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So we have our answer to the problem.
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The length of the perpendicular between the point one, nine and the straight line negative five 𝑥 plus 12𝑦 plus 13 equals 0 is 116 over 13.
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Find the length of the perpendicular drawn from the point 𝐴 negative one, negative seven to the straight line passing through the points 𝐵 six, negative four and 𝐶 nine, negative five.
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So the first thing we need is we need to know the equation of the line that joins the points 𝐵 and 𝐶.
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So I’ll find this equation using the point slope method.
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𝑦 minus 𝑦 one equals 𝑚 𝑥 minus 𝑥 one.
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First, we’ll find the value of 𝑚, the slope of the line, using 𝑦 two minus 𝑦 one over 𝑥 two minus 𝑥 one.
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It doesn’t matter which way round I allocate 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two, so I’ve chosen to allocate it in this order.
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It gives me negative five minus negative four over nine minus six.
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This gives me a slope of negative a third for this line.
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Next, I’m gonna substitute one of these points into the equation of the line.
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And again, it doesn’t matter which I choose.
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I’ve chosen to substitute the coordinates of 𝐵.
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So I’ve substituted six for 𝑥 one and negative four for 𝑦 one.
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This simplifies to give 𝑦 plus four is equal to negative a third 𝑥 plus two.
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And finally, if I multiply the whole equation through by three and group all the terms together on the left-hand side, then it gives me the equation of the line, which is 𝑥 plus three 𝑦 plus six is equal to zero.
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So we could answer this problem using the formula that’s on the screen, which tells us that the shortest distance or the perpendicular distance between the line 𝑎𝑥 plus 𝑏𝑦 plus 𝑐 equals zero and the point 𝑥 one, 𝑦 one is given by the modulus of 𝑎𝑥 one plus 𝑏𝑦 one plus 𝑐 all over the square root of 𝑎 squared plus 𝑏 squared.
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However, I’m going to work through the maths and the logic behind that formula in this question.
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So I’ve just drawn a sketch to help visualise the situation.
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I’ve got the line 𝑥 plus three 𝑦 plus six equals zero and the point 𝐴 with coordinates negative one, negative seven.
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And it’s this distance 𝑙 that I’m looking to find.
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The first thing I’m going to do is find the equation of the line that connects 𝐴 to the line 𝑥 plus three 𝑦 plus six equals zero.
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There are now two things about this line.
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I know the coordinates of one point on the line, negative one, negative seven.
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And I know it’s perpendicular to the line 𝑥 plus three 𝑦 plus six equals zero.
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So this is enough information to enable me to find its equation.
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As it’s perpendicular to the first line, I can use the fact that the slopes of perpendicular lines are the negative reciprocals of each other.
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They multiply to negative one.
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Rearranging the equation of the first line gives me 𝑦 equals negative a third 𝑥 minus two.
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So I can see that the slope of the first line is negative a third.
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This means that the slope of a perpendicular line must be three.
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So I’m going to find the equation of this line using the point slope form.
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𝑦 minus 𝑦 one equals 𝑚 𝑥 minus 𝑥 one.
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I now know that 𝑚 is three.
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And I know that the point I’m going to use, the 𝑥 one, 𝑦 one, is the point with coordinates negative one, negative seven.
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So this gives me 𝑦 minus negative seven is equal to three 𝑥 minus negative one.
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This can be simplified in a couple of lines of algebra to give the equation of the line 𝑦 equals three 𝑥 minus four.
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So now I have the equations of both lines.
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There are two more stages in the method that I’m using here.
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The next stage is to find the coordinates of the point of intersection of the two lines, the point that I’ve labelled in orange on the diagram.
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To do this, I need to solve the equation of the two lines simultaneously.
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I’m going to do this using substitution, by substituting the expression for 𝑦 from equation two into equation one.
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So this gives me 𝑥 plus three lots of three 𝑥 minus four plus six is equal to zero.
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Now there are a couple of lines of algebra to work through in order to solve that equation for 𝑥.
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And I’ll leave you to fill those in yourself.
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But if you do them correctly, it will lead to 𝑥 is equal to three-fifths.
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Now we need to find the value of 𝑦.
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So I’m gonna substitute 𝑥 is equal to three-fifths into equation two.
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So we have 𝑦 is equal to three multiplied by three-fifths minus four.
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And this gives us a value of negative 11 over five.
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So now we know the coordinates of the point of intersection of the two lines.
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And there’s only one step left in our method.
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We need to find the distance between point 𝐴 and this point of intersection that we’ve just calculated.
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So in order to do this, we can use the distance formula, which tells us that the distance 𝑙 is equal to the square root of 𝑥 two minus 𝑥 one all squared plus 𝑦 two minus 𝑦 one all squared.
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Where 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two represent the coordinates of the two points that we’re looking to find the distance between.
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So substituting the values for 𝑥 one, 𝑥 two, 𝑦 one, and 𝑦 two gives me this calculation here for 𝑙.
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Now if you evaluate this on a calculator or if you work through the arithmetic yourself, it leads you to the square root of 640 over 25.
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But this can be simplified.
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So we have a final simplified answer of eight root 10 over five length units.
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You can confirm that this is the same answer we would’ve achieved if we had used that standard formula earlier within the question.
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If the length of the perpendicular drawn from the point negative five 𝑦 to the straight line negative 15𝑥 plus eight 𝑦 minus five equals zero is 10 length units, find all the possible values of 𝑦.
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So this question is asking us about the length of the perpendicular from a point to a straight line.
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So we need to recall the standard formula for calculating this.
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The formula is this, that the length of the perpendicular from the point 𝑥 one, 𝑦 one to the line 𝑎𝑥 plus 𝑏𝑦 plus 𝑐 equals zero is given by the modulus of 𝑎𝑥 one plus 𝑏𝑦 one plus 𝑐, all divided by the square root of 𝑎 squared plus 𝑏 squared.
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We’re also told that, within this particular question, this length is 10 length units.
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So let’s compare the general information with the specific values in this question.
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If I look at the equation of the line first of all, this tells me that 𝑎 is equal to negative 15, 𝑏 is equal to eight, and 𝑐 is equal to negative five.
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Now if I look at the coordinates of the points that we’re interested in, this tells me that 𝑥 one is equal to negative five and 𝑦 one has just been given the general coordinate 𝑦.
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I’m looking to find the possible values of 𝑦.
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So I need to use this information to set up and then solve an equation.
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So substituting those five values into the relevant places in the formula and remembering that I’m told that it’s equal to 10 gives me this equation here.
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Evaluating the numeric parts of this tells me that I have the modulus of 70 plus eight 𝑦 divided by 17 is equal to 10.
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And multiplying by 17, I have that the modulus of 70 plus eight 𝑦 is equal to 170.
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Now let’s look at what this modulus means.
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It means the absolute value of 70 plus eight 𝑦.
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So 70 plus eight 𝑦 is either equal to 170 or it could be equal to negative 170.
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This means that I have two linear equations to solve, leading to two possible values of 𝑦.
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The first equation gives eight 𝑦 is equal to 100, and then 𝑦 equals 100 over eight, which simplifies to 25 over two.
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The second equation gives eight 𝑦 equals negative 240.
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And then dividing by eight gives 𝑦 equals negative 30.
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So in this question, there are two possible values of 𝑦.
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𝑦 is either equal to 25 over two or it’s equal to negative 30.
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In summary then, we’ve seen that the shortest distance between a point and a line is the perpendicular distance.
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We’ve seen the formula for calculating this distance.
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𝑙 is equal to the modulus of 𝑎𝑥 one plus 𝑏𝑦 one plus 𝑐, all divided by the square root of 𝑎 squared plus 𝑏 squared.
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We’ve also seen how to take a more first-principles-like approach to answering such a problem by solving a system of linear equations.