Distances, Feet of Perpendiculars and Images

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. You stand at the point four comma three, and along the line x plus y equals four, there is a long, flat mirror. Where exactly is your reflection? Today we learn the distance of a point from a line, the distance between parallel lines, and the foot of the perpendicular and the image of a point. These come up often in JEE coordinate geometry. So, let's begin.

0:29Distance of a point from a line

So, here is the picture, on the usual x and y axes. Take a point P, x one comma y one, in round brackets; the small ones are subscripts, which just number the point. And take the dark line, a times x, plus b times y, plus c, equals zero. P prime, the point at the bottom with a little dash, we will meet later. Okay? Now let me ask you this. If you walk from P to this line, which path is the shortest? Yes, exactly, the perpendicular! It meets the line at M, and the little square there marks the right angle. M is called the foot of the perpendicular, and the length P M, marked d, is the distance of P from the line. Clear? And here is the formula. d equals the modulus of, a times x one, plus b times y one, plus c, over the square root of, a squared plus b squared. The two straight bars are the modulus sign, the long tick is the square root, and the small two up top means squared. So, put the point into the left side, take the modulus, and divide. Got it?

1:40Why the modulus? Which side of the line?

Now, look at the formula once more. Why do we need the modulus at all? Yes, it is all about the sign! Without the modulus, a times x one, plus b times y one, plus c, can come out negative, and a distance is never negative, right? It is positive on one side of the line, negative on the other, and zero on the line itself. Okay? And that gives a lovely test from the course. Two points lie on the same side of the line if and only if a times x, plus b times y, plus c, has the same sign at both; the double arrow means if and only if. Opposite signs mean the line passes between them. With me?

2:22Distance between two parallel lines

Next, two parallel lines: same a, same b, and only the constant changes, c sub one and c sub two. Same a and b means the same direction, so the lines are parallel. Right? The distance between them is the modulus of, c one minus c two, over root of a squared plus b squared. It looks really simple, but there is a trap hiding here, okay? Look at these two lines from the course: three x plus four y equals five, and six x plus eight y equals thirty. Can we just take c one as minus five, and c two as minus thirty? No, no, careful, that is exactly the trap! The formula needs the same a and b in both lines. So divide the second line by two; the arrow means, which gives, three x plus four y equals fifteen. Clear? Now c one is minus five and c two is minus fifteen. Minus five minus, minus fifteen, gives ten, and root of three squared plus four squared is five. Ten over five gives two units. Got it?

3:35The foot of the perpendicular and the image

Now the third result, the one for our mirror. x minus x one, over a, equals y minus y one, over b, equals minus lambda, times, a times x one, plus b times y one, plus c, over a squared plus b squared. Lambda, the Greek letter, is a number we choose. The first two fractions keep us on the perpendicular through P, okay? Put lambda equals one, and the point x comma y you get is the foot M, where the perpendicular meets the line. Right? Now, think of the line as a mirror. The image of P is P prime, written with the little dash called a prime. Which value of lambda gives the image? Yes, exactly, lambda equals two! The image is as far behind the mirror as P is in front, so we go twice as far as the foot, and M sits right in the middle of P and P prime, as the figure shows. And keep the minus sign in front of lambda; drop it, and you walk away from the line.

4:39Example 1: distance of (3, -2) from 3x + 4y = 11

Clear? Alright, the first one from the course. Find the distance of the point three comma minus two from the line three x plus four y equals eleven. Pause the video and try it first. Done? Okay, let's check. First, what is c here? Many students write eleven. Is that right? No, no, careful, the formula needs zero on the right. So the line is three x plus four y minus eleven equals zero, with a equals three, b equals four, and c equals minus eleven. Right? Now put the point in. On top: three times three, plus four times minus two, minus eleven; the brackets just mean times. At the bottom: root of three squared plus four squared. Okay? Nine minus eight minus eleven is minus ten, and its modulus is ten. Root of twenty five is five. So ten over five gives two. The distance is two units, and that's it!

5:41Example 2: foot and image of (4, 3) in x + y = 4

Now, remember our mirror puzzle? Let's solve it. Find the foot of the perpendicular and the image of the point four comma three in the line x plus y equals four. The figure on screen is exactly this problem. Pause the video and try it first. Done? Okay, let's check. Put zero on the right: x plus y minus four equals zero, so a equals one, b equals one, and c equals minus four. Right? Now the common piece: a times x one, plus b times y one, plus c, over a squared plus b squared. That is four plus three minus four, over one plus one, which is three over two. With me? For the foot, lambda is one, and since a and b are both one, we subtract three over two from each coordinate of the point. Four minus three over two is two point five, and three minus three over two is one point five. So the foot is M, two point five comma one point five. Clear? Now the image. Tell me, what changes when we go from the foot to the image? Yes, exactly, only lambda changes! With lambda equal to two, we subtract twice three over two, which is three, from each coordinate. So P prime is four minus three, comma, three minus three, which is one comma zero. Puzzle solved: your reflection is at one comma zero, right on the x axis! And always check your answer. The midpoint of P and P prime is four plus one, over two, comma, three plus zero, over two, which is two point five comma one point five. That is exactly M, a perfect check!

7:30Quick revision: distance, parallel lines, foot and image

So, quick revision. Point to line: the modulus of, a times x one, plus b times y one, plus c, over root of a squared plus b squared. Parallel lines: the modulus of, c one minus c two, over the same root, after making a and b the same. Foot: lambda is one; image: lambda is two, twice as far along the perpendicular. And same side means the same sign of a times x, plus b times y, plus c. Now go and try the two practice problems on this page. All the best!