Mixed Problems: Set 1 (Points and Lines)

Transcript

0:00Introduction

Hello everyone! Okay, a quick challenge before we start. Two lines cross, and a question asks for the bisector of the angle that contains the origin. The formula gives two answers, one with a plus sign, one with a minus sign. So which one do you pick? Many students pick plus straight away, and in today's harder problem that gives the wrong line. This is Set one of our mixed problems, points and lines, and such questions come up often in JEE Main. So, two quick tools first, then two problems from the set. Let's begin!

0:37Tool 1: the foot and the image of a point

So, here is the course's picture. The point P is x one comma y one; the small ones are subscripts. The dark line is a times x, plus b times y, plus c, equals zero. The orange perpendicular meets it at M, the foot, and the small square marks the right angle. The length P M is d, the distance: the modulus of, a times x one, plus b times y one, plus c, over root of a squared plus b squared. The two straight bars are the modulus sign, and the long tick is the square root. And the point P prime at the bottom, with the little dash called a prime, we will meet in a minute. Okay? Now one formula gives the foot and the image together. 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 is a Greek letter, a number we choose, and the small two up top means squared. Right? Put lambda equals one, and the point x comma y you get is M, the foot. Clear? Now, P prime is the image of P, as if the line were a mirror. Let me ask you this: which value of lambda gives the image? Yes, exactly, lambda equals two! The image is as far behind the line as P is in front of it, so it is twice as far from P as the foot. With me? And here is a topper's ten second check. M is the mid-point of P and P prime, so the mid-point of P and your answer must lie on the line. Got it?

2:14Tool 2: the angle bisectors, and which sign

Now tool two, for two lines, L one and L two. L one is a one times x, plus b one times y, plus c one, equals zero, and L two has twos in place of the ones. Okay? Let me ask you this: what is the same for every point on a bisector? Yes, exactly, it is equally far from both lines! Equate the two distances and open the modulus signs: that gives plus or minus, the plus sign over a minus sign on screen. So the bisectors are: a one x plus b one y plus c one, over root of a one squared plus b one squared, equals plus or minus the same thing for the second line. Right? Now comes the question from our challenge. Which sign gives the bisector of the angle that contains the origin? If you said the plus sign, you are only half right! Here is the rule from the course. First make c one and c two the same sign, by multiplying one equation by minus one if needed. Then the plus sign bisects the angle containing the origin. This works because at the origin the two expressions are just c one and c two, now of the same sign. Clear? Here is one more rule from the course. With the constants of the same sign, if a one times a two, plus b one times b two, is negative, this plus bisector is the acute one. The small sign that opens to the left means less than.

3:40Example 1: image of (1, 2) in the line x - y + 3 = 0

Got it? Alright, the first one is problem five of Set one, marked easy. Find the image of the point one comma two in the line x minus y plus three equals zero. So P is one comma two, and we want P prime from the picture. Start your clock, pause the video, and try it first. Done? Okay, let's check. Compare with a times x, plus b times y, plus c, equals zero: so a is one, b is minus one, and c is three. Put the point in: one minus two plus three gives two. Okay? We want the image, so lambda is two. x minus one, over one, equals y minus two, over minus one, equals minus two, times two over two; the dot means times. Here a squared plus b squared is two, so everything equals minus two. Right? So x minus one is minus two. And y minus two is also minus two, so y is zero, right? No, no, wait, look at the bottom of that fraction! y minus two is over minus one, so y minus two equals minus one times minus two, which is plus two. So x is one minus two, which is minus one, and y is two plus two, which is four. Clear? So the image is minus one comma four. Now our ten second check: the mid-point of one comma two and minus one comma four is zero comma three. Put it in the line: zero minus three plus three is zero. It lies on the line, so our answer passes the check. This check would also have caught that sign slip. With me?

5:20Example 2: bisector of the angle containing the origin

Now a harder one, problem seven of Set one, marked medium, and it is our opening challenge. Find the bisector of the angle between three x minus four y plus seven equals zero, and twelve x plus five y minus two equals zero, that contains the origin. Here is a hint: check the constants first. Pause the video and try it. Done? Okay, let's check. The origin's angle means the plus sign, so use both equations just as written, right? Oh no, careful, look at the constants! Plus seven and minus two have opposite signs. So first, multiply the second equation by minus one, and it becomes minus twelve x minus five y plus two equals zero; the arrow means which gives. Skip this, and the plus sign gives the other bisector, for the angle without the origin. Got it? Next, we need the two roots. Root of three squared plus four squared is five, and root of twelve squared plus five squared is thirteen. Okay? Now we take the plus sign, for the origin's angle. Three x minus four y plus seven, over five, equals minus twelve x minus five y plus two, over thirteen. With me? Now let us cross multiply. Thirteen times the left side gives thirty nine x minus fifty two y plus ninety one. Five times the right side gives minus sixty x minus twenty five y plus ten. Bring everything to the left: ninety nine x minus twenty seven y plus eighty one equals zero. Right? Every number here is a multiple of nine, so divide by nine. Eleven x minus three y plus nine equals zero. That's it, and our challenge is solved. Clear? And here is a bonus from tool two. With the constants of the same sign, a one times a two, plus b one times b two, is three times minus twelve, plus minus four times minus five. Minus thirty six plus twenty gives minus sixteen, which is less than zero. So this bisector is also the acute angle bisector. Got it?

7:27Quick revision: foot and image, same-sign rule, sign traps

So, quick revision. Foot, lambda is one; image, lambda is two, and the mid-point of P and its image lies on the line. Bisectors: make the constants the same sign first, then the plus sign gives the origin's angle. And read every sign, like a minus one under y minus y one, or a constant like minus two. Now start your clock, try the other problems of Set one, and the practice problem on this page. All the best!