Angles, Intersections and Families of Lines

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take a line whose equation has a k inside it. Put k equals zero, and you get one line; put k equals one, and you get a different line, and so on, infinitely many lines. And still, every single one of them passes through the same point. How is that possible, and how would you find that point? In this video, we look at the angle between two lines, three lines through one point, and families of lines. This comes up often in JEE coordinate geometry, so let's begin.

0:36The angle between two lines

So, two lines cross, with slopes m one and m two; the small one and two are subscripts that just number the lines. Now look, two crossing lines make two angles. Theta, the Greek letter, is the first angle, and the other is one hundred and eighty degrees minus theta, with the little circle as the degree sign. Okay? Here is the formula. tan theta equals the modulus of, m one minus m two, over one plus m one times m two. The two straight bars around the fraction are the modulus sign. With the modulus, this theta is always the acute angle, less than ninety degrees. Right? Let me ask you this. Why do we need the modulus at all? What goes wrong without it? Yes, exactly, it is all about the sign! Without the modulus, the fraction can come out negative. And the tan of one hundred and eighty degrees minus theta is minus tan theta. So a negative value belongs to the obtuse angle, not the acute one. The modulus fixes that. Clear? Now, what if the denominator, one plus m one times m two, is zero? Then m one times m two equals minus one, and tan theta is not defined, so theta is ninety degrees. The lines are perpendicular, exactly the rule from the last concept. With me? But this formula needs both slopes to exist. Now, a vertical line has no slope, so neither this formula nor the product m one times m two works for it. Still, a vertical line is perpendicular to every horizontal line. Got it?

2:15Example 1: angle between y = 2x + 1 and y = -3x + 2

Alright, let's try one from the course. Find the angle between the lines y equals two x plus one, and y equals minus three x plus two. Pause the video and try it first. Done? Okay, let's check. Both lines are in the form y equals m x plus c, so the slope is the number in front of x. So m one is two, and m two is minus three. Right? Now into the formula. The top is two minus, minus three, which is five. The bottom is one plus two times minus three, so one minus six, which is minus five. Five over minus five gives minus one, and its modulus is one. Clear? Now here is the trap. Suppose you forgot the modulus. Then tan theta would be minus one. Which angle would you have written down? Exactly, one hundred and thirty five degrees, the obtuse angle, and that is the wrong answer here. With the modulus, tan theta equals one, so theta is forty five degrees, the acute angle, and that is the answer.

3:23Three lines through one point

Now, let's move to the second idea. Three lines are called concurrent when all three pass through one common point, okay? Write the lines in general form: a sub i times x, plus b sub i times y, plus c sub i, equals zero, where i takes the values one, two and three. Right? Then put the coefficients in a determinant, one row for each line: its a, its b and its c. The straight bars on the sides make it a determinant. The lines are concurrent if and only if it is zero; that is what the double arrow means. Clear? But wait, can you think of three lines where this determinant is zero, yet there is no common point? Yes, exactly, three parallel lines! For parallel lines, the a and b coefficients are in the same ratio, so the determinant is zero, but the lines never meet. So the course adds this condition: the lines are not all parallel. Okay? And in practice, the course examples use a quicker way. Find where two of the lines meet, then put that point into the third. If it fits, they are concurrent. Got it?

4:33Families of lines

Now the third idea, a beautiful one. Take two lines, L one equals zero and L two equals zero, where L one and L two stand for the left sides of their equations. Say they meet at a point, P, okay? Now form L one plus lambda times L two, equals zero. Lambda, the Greek letter, can be any number. For every value of lambda, this is a straight line through the point P, right? We call it the family of lines through the point of intersection. Let me ask you this. Why must every one of these lines pass through the point P, whatever lambda is? Yes, exactly, because at the point P, both L one and L two are zero. The arrow means implies: the left side is zero plus lambda times zero, the dot meaning times, so it is zero for every lambda. With me? Now turn this around. If a line's equation depends linearly on a parameter k, meaning k appears only to the first power, the line usually passes through a fixed point, whatever k is. One exception: if all the lines are parallel, like y equals x plus k, there is no fixed point at all. Our puzzle is the usual kind. Okay? So here is the method. Collect the terms with k and take k common: k times one bracket, plus another bracket, equals zero. The three dots stand for whatever is inside. Then set both brackets equal to zero, and solve them together. Clear?

6:08Example 2: fixed point of (2k+1)x + (k-1)y - (4k-1) = 0

Now, let's finally solve our puzzle. Show that two k plus one, times x, plus k minus one, times y, minus the bracket four k minus one, equals zero, passes through a fixed point for every k, and find that point. Pause the video and try it first. Done? Okay, let's check. First, open all the brackets: two k x plus x, plus k y minus y. Oh no, careful with the minus sign in front of the last bracket! Minus, four k minus one, becomes minus four k plus one. Right? Now tell me, which terms carry a k? Find them, and take k common. Yes, exactly: two k x, k y, and minus four k. So we get k times, two x plus y minus four, plus the rest, x minus y plus one, equals zero. Got it? For a point on every line of this family, both brackets must be zero. So two x plus y equals four, and x minus y equals minus one. Clear? Add the two equations, and the y terms cancel: three x equals three. So x is one, and then y is two. The fixed point is one comma two. Puzzle solved: infinitely many lines, all through the point one comma two! And always check your answer. Put one comma two into the collected form: k times, two plus two minus four, plus, one minus two plus one. Both brackets are zero, so it holds for every k, a perfect check!

7:50Quick revision: angle, concurrency and families of lines

So, quick revision. The angle between two lines: tan theta equals the modulus of, m one minus m two, over one plus m one times m two, and that is the acute angle. Three lines are concurrent when the determinant is zero and they are not all parallel. And L one plus lambda L two equals zero gives the lines through the intersection; an equation linear in k means a fixed point, unless all the lines are parallel. Now go and try the two practice problems on this page. All the best!