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AS Level Mathematics 9709 | Coordinate Geometry 1 | Distance Formula & Ratio | Raja Shafqat — Transcript

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0:08Assalamu Alaikum beta, today we are
0:10finally going to start the third topic
0:13of our AS Maths, InshaAllah, and the
0:15name of this third topic is Coordinate
0:17Geometry. Okay? Coordinate Geometry. I
0:20will give you a brief introduction to
0:21it. There are basically two concepts in
0:23Coordinate Geometry. One concept is
0:25called Basic Coordinate Geometry. Basic
0:29Coordinate Geometry. Well, in this
0:31basic coordinate geometry, you will
0:33also see O-Level Math. Clear? You will
0:36also find O-Level Math in basic
0:37coordinate geometry. Like distance,
0:39midpoint, gradient, and the equation of
0:41a straight line. Plus, the new
0:43coordinate geometry is read as
0:45Coordinate Geometry of Circles. This is
0:48something new. Coordinate Geometry of
0:52Circles. This is basically a new thing,
0:55Coordinate Geometry of Circles. It was
0:59added to AS Maths in May/June 2020.
1:02Clear, beta? It was added to AS Maths
1:05in May/June 2020. This is a slightly
1:08tricky part. Okay? What is required for
1:11this? For this, you will first need the
1:13entire basic coordinate geometry. Basic
1:15coordinate geometry will be required.
1:17Plus, after that, you will need the
1:20properties of a circle. Geometrical
1:23properties of a circle will also be
1:25required. Clear, beta? Alright, so now
1:27let's finally start. Obviously,
1:30whenever we talk about coordinate
1:32geometry, all the work in coordinate
1:34geometry is obviously about coordinates
1:36. Okay? We must have studied in O-Level
1:39Math that we represent a point like
1:41this in coordinate form. What is the
1:43name of this x? What does x mean? x is
1:45called horizontal distance in the
1:47language of math. x represents
1:49horizontal distance in the language of
1:52math. Meaning, how much you move
1:54horizontally, that is, in the direction
1:56of the x-axis from the origin. Meaning,
1:59how many units you move left or how
2:01many units you move right. We call this
2:04x-coordinate the horizontal distance.
2:06So, what does the x-coordinate show?
2:07Horizontal distance. And this
2:09y-coordinate, what does the
2:11y-coordinate show? The y-coordinate
2:13shows vertical distance. The
2:16y-coordinate shows the vertical
2:18distance. What does vertical distance
2:21mean? How much you move up or down from
2:24the origin. So, this distance up or
2:26down is what we call vertical distance
2:28in the language of math. Vertical
2:31distance. So, it means that whenever
2:33you show a point x, y in math, it means
2:36you will check how much you are moving
2:38left or right from the origin and how
2:40much you are moving up or down from the
2:42origin. Okay, beta? These axes that we
2:46use in O-Level Math, we call these axes
2:49coordinate axes. Obviously, this is
2:51your + x. This is your-x. This is your
2:54+ y and this is-y, which we study in
2:56basic O-Level Math. So, what do we call
2:58these axes? We call these axes
3:01coordinate axes. What word or term do
3:05we use in Math for the x and y axes? We
3:07use the word or term coordinate axes.
3:10For this, we use the Math term
3:13coordinate axes. Coordinate axes. What
3:18is the Math word we use for this?
3:20Coordinate axes. In the language of
3:22Math, the x-axis and y-axis are called
3:24coordinate axes. Or you could say, in
3:27simple language, we also use this term.
3:30x and y axes. What is the meaning of
3:33coordinate axes? x and y axes. Clear,
3:36kiddo? Normally, if a point is given as
3:402, 3, we say we move from the origin 2
3:41units right and 3 units up, and we mark
3:43the point. Or if we are asked to mark
3:47this point, how do we mark it? 2 units
3:50left, 3 units up. You marked the point.
3:53Or if your coordinate is given as 2 and
3:56-3, it means from the origin you will
3:59move 2 units right and 3 units down. Or
4:02if your point is given as -2 and -3,
4:05you will move 2 units left from the
4:07origin and 3 units down. By the way,
4:10this is the same basic Math that we
4:11study in O-Level Math. Clear, kiddo? If
4:14you have a point given as 0, 5, what
4:17does 0, 5 mean? No right or left. So,
4:21where will you move straight away? You
4:23will move 5 units upward. So, we are in
4:25coordinate geometry. The whole game is
4:27about coordinates. So, what do we call
4:29this position in Math? 'x' coordinate.
4:32'x' coordinate means left or right
4:34distance, and the second position we
4:36call the 'y' coordinate. In O-Level
4:38Math, we call the second position the '
4:40y' coordinate. 'y' coordinate means
4:42vertical distance, that is, how much to
4:44move 'up' or 'down' from the origin.
4:47Okay, let's start with a small formula
4:49first. Before this, we must have
4:51studied a right-angled triangle in
4:53O-Level Math. We also studied
4:55right-angled triangles in O-Level Math
4:57or circular measurement. When we talk
4:59about any right-angled triangle, any
5:01right-angled triangle at all, we say
5:02suppose this is your base, this is the
5:04opposite, and this is the hypotenuse.
5:07So, we use a small formula in O-Level
5:10Math, c² = a² + b², which in the
5:13language of math we call the
5:15Pythagorean theorem. What is this
5:18formula? Pythagorean theorem. All right
5:21? Now look, let's start the first
5:23concept. The first formula in AS Math
5:25is the distance formula. This was in
5:27O-Level Math too. By the way. Another
5:29word is also used in math for the
5:30distance formula. Length of a line
5:32segment. Basically, in math, we have
5:35studied in O-Level Math that a line
5:37itself is always infinite. So, if you
5:39are interested in working between any
5:41two particular points on a line. That
5:44is, if you need to find the distance or
5:46length between these two particular
5:48points, what word do we use in math for
5:50that? We use the term length of a line
5:52segment. So, by the way, what is a line
5:55segment? A line segment is called a
5:57part of a line. Part of a line. So, if
6:02I tell you that, hey, I have this line
6:04and there are these two points on it.
6:06By the way. These are two points. I
6:08will read the coordinates of the first
6:10point as x1 and y1. This point was also
6:13in O-Level Math. x1 and y1 and there is
6:15another point. I will read the
6:17coordinates of point B as. x2 and y2.
6:20These are two different points. We read
6:22their coordinates as 1-1 and 2-2 to
6:25differentiate them. Now look, when we
6:27talked about the distance formula in
6:29O-Level Math, the distance formula
6:31means that a right-angled triangle will
6:33be formed between these two points. A
6:35right-angled triangle will be formed.
6:37So that means when you move from this
6:39point to this point, what change will
6:41occur horizontally? You will move
6:43horizontally and then you will move
6:45some vertical distance. So that means
6:48if I ask you what the horizontal
6:49distance is. Then the horizontal
6:51distance is x2-x1, that is, the
6:53distance between x-coordinates, the
6:55difference of x-coordinates. So what
6:58does this basically show? It shows the
7:01difference of x-coordinates. Difference
7:04of x-coordinates. And similarly, the
7:06vertical difference, the difference
7:08between y-coordinates, what do we call
7:10that? We read that as the opposite.
7:12Opposite. So, what is your target? Your
7:15target is the distance between two
7:16points. So, in O-Level Math, we used to
7:19show distance like this. We used this
7:22formula for the distance modulus AB, or
7:24we would say distance AB, meaning the
7:26length of a line segment between points
7:28A and B. So, what did we learn in
7:30O-Level Math? That the distance formula
7:33essentially means the Pythagorean
7:35theorem. So basically, if you need the
7:37distance, what are you indirectly
7:39looking for? Indirectly, you are
7:41looking for the hypotenuse. So, what
7:42does that mean the formula for the
7:44hypotenuse is? The hypotenuse formula
7:47is the base squared plus the
7:49perpendicular squared. That is why we
7:52must have used this formula in O-Level
7:54Math. In mathematical language, we read
7:56this formula as the distance formula. (
7:59x2-x1) ² + (y2-y1) ², so technically,
8:03what is this? This is the x-difference.
8:06And technically, what is this? This is
8:09the y-difference between your two
8:11points. Meaning the horizontal gap and
8:13the vertical gap. We used to read this
8:16formula in O-Level Math. The distance
8:19formula: (x2-x1) ² + (y2-y1) ². Is
8:25that clear, son? And you will also
8:27notice one thing: when is the distance
8:29formula always applied? When a
8:31right-angled triangle is formed. Let me
8:32guide you through the details. Look,
8:34let me give you a small example. Let me
8:36give you a small example. Suppose you
8:38have a line, and there are two
8:40different points on this line. I am
8:42just giving a random example. Suppose
8:44you have a point A. I will consider the
8:46coordinates of point A to be 3, 4. Here
8:49you go. I have considered this point 3,
8:514; this is a point on the line. Its
8:54coordinates are 3, 4, and I will
8:55consider another point just as randomly
8:58. Suppose the coordinates of this point
9:00are 6, 8. Okay? I just marked two
9:02points randomly. My target is to find
9:05the distance between A and B. So, one
9:08option is that if you want to apply the
9:11distance formula, what does it mean to
9:13apply the distance formula here? Look,
9:16if my target is the distance formula.
9:18Then what is this point? x1, y1, and
9:20how will I read this point? x2, y2. x1,
9:23y1, x2, y2. So, one option is that I
9:26apply the formula straight away. How
9:29will I apply the formula? The formula
9:32will be (x2-x1) ², the formula will be
9:36like this: (6-3) ². Here is (x2-x1) ²
9:40. And what will be added?(y2-y1) ². (
9:46x2-x1) squared, like we used to apply
9:48in O-Level Math, and (y2-y1) squared.
9:50What is this? y2, and what is this? y1.
9:53Is that clear, son? So we applied the
9:56formula x2-x 1² and y2-y 1², see, we
9:59applied the formula, is it clear, kids?
10:02We applied this formula straight away,
10:05here is your formula x2-x 1² and y2-y
10:091², and when I simplify this further,
10:12what result will be derived? The result
10:15will be derived. What will 6-3 become?
10:173², here it is. What did 6-3 become?
10:206-3 will be 3 squared, and what will
10:238-4 become? 4 squared. So this means
10:26you have simplified it. When you
10:28simplified it, what answer was derived?
10:3125, and what is the root of 25? The
10:34root of 25 will be 5 units. What will
10:38the root of 25 become? 5 units.
10:41Otherwise, you can deal with the same
10:43question by asking what this
10:45technically means. Technically, it
10:47means that when you require the
10:49distance, the distance means a
10:50right-angled triangle is being formed
10:52between these two points. A
10:54right-angled triangle is being formed
10:56between these two points. Now look, its
10:58x-coordinate is three. Its x-coordinate
11:00is six. So what is the horizontal gap
11:02between three and six? Three. The
11:04y-coordinates are four and eight. What
11:06is the vertical gap between them? Four.
11:08So that means this gap here is three.
11:11The horizontal distance is three. And
11:14what is the vertical distance? The
11:15vertical distance is four. So now this
11:17is three and this is four. So what is
11:19your target? Basically, your target is
11:22the hypotenuse. So if I apply the
11:24distance formula directly, I can also
11:26say directly, what is the distance
11:27formula? The distance formula is 3². +
11:324². So when I simplify this, what
11:36result will ultimately be derived? The
11:38result that will ultimately be derived
11:40is 25 units. So what is the final
11:42answer derived for 25 units? 5 units.
11:46What will be the square root of 25? 5
11:50units. So, the gap between these two
11:53points, the length of this segment, is
11:56what? 5 units is the length of this
11:59segment. So what basically is the
12:01distance formula? The distance formula
12:04is basically another name for the
12:05Pythagorean theorem. The distance
12:07formula is basically another name for
12:10the Pythagorean theorem. Clear, kids?
12:12So we studied this in O-Level Math. We
12:14cover this formula in O-Level Math.
12:16Distance formula. What else do we call
12:19it? Length of a line segment. Length of
12:22a line segment. I am explaining one
12:24more small point. So I used it a little
12:26while ago. I used a word that the
12:29distance formula applies when any point
12:32is on a tilted line. Let me give you a
12:34small example to make the distance
12:36formula a bit clearer. Look, let me
12:38give you a small example. Suppose you
12:41have this condition. Suppose you have
12:44this y-axis. And suppose you have this
12:47point. Let's label the coordinates of
12:49this point. Uh, let's say the
12:51coordinates of this point are 0, 4.
12:55This is a point on the y-axis. And
12:58suppose you have this point. You have
13:00this point here. Its coordinates are 0,
13:021, right here. And your target is the
13:05distance between A and B. Your target
13:07is the distance between A and B. You
13:09require the distance between A and B.
13:12So how will the distance AB be found?
13:14Now, what does common sense say? You
13:16already know that both these points are
13:18on the vertical axis. There is no
13:20change in X for them. So, there is a
13:22vertical change. What is the vertical
13:24gap between one and four? The vertical
13:26gap between one and four, you already
13:28know what that vertical gap is. Three
13:31units. That vertical gap is three units
13:33. So common sense says that in this
13:36condition, you should not apply the
13:38distance formula. Because the
13:40Pythagorean theorem is not being formed
13:42here. Even then, if you directly apply
13:44the distance formula, when you treat it
13:47as the distance formula, the answer
13:49will still be the same. But this would
13:51be a weak method in this condition.
13:54Read this question. These coordinates
13:56are x1, y1 and this point is x2, y2. So
13:59if you require the distance formula,
14:01what will you do? x2-x1 squared, plus
14:06what will it be? Plus will be y2-y1
14:10squared, so when you simplify this
14:12further, what answer will be derived?
14:15The answer derived will be 0 + 9, 0 + 9
14:18, and what will the square root of 9 be
14:21found? The square root of 9 will be
14:24found to be 3 units. So common sense
14:26says that we should not apply this
14:28method in this condition. Common sense
14:31says that this is a slow method. So
14:33that is why sometimes it also happens
14:34in coordinate geometry that both your
14:36points are on a vertical line. And you
14:38have to find the difference or distance
14:40between those two vertical lines. So
14:42math says that in this condition, you
14:45should not apply the Pythagorean
14:47theorem. Because it is a weak method.
14:49We wasted time for a basic thing. Clear
14:52, kid? We wasted time for a basic thing
14:55. What if this exact condition appeared
14:57in the exam? Look, suppose you have a
14:59situation where a point is given on the
15:02y-axis. This is the point. Its
15:03coordinates are 0, 3. This is the point
15:06. Its coordinates are 0, 3, and here is
15:08another point. Its coordinates are 0,
15:12-2. So, our target is to find the
15:15distance between A and B. Clear?
15:19Distance between A and B; now look,
15:21what does common sense say if you need
15:23the distance between A and B, or you
15:25can simply show it like this as AB?
15:27Just see what the vertical gap is
15:29between these two. If I ask you what
15:31this entire vertical gap is, how would
15:34we calculate this whole vertical gap?
15:37Look closely, child; you already know
15:40what the distance above the x-axis is.
15:43The distance above the x-axis is 3
15:45units. We know that length is always
15:47positive, and if I ask you, what is
15:49this distance? The distance below the
15:51x-axis. So, what does the minus sign
15:53actually show? The locus or the
15:55location of where your point is. Above
15:58the x-axis or below the x-axis. So,
16:00what will this length obviously be?
16:02This length will obviously be two. Now,
16:04add these two distances together. So,
16:06what length have you found? The length
16:08found is 5 units. So, obviously, in
16:11this basic condition, we should not
16:13apply the Pythagorean theorem or the
16:15distance formula. Look at another
16:18example. Suppose you have this
16:20situation, suppose you have this
16:23situation. I marked a point. The
16:25coordinates of this point are 5, 0;
16:28this point is already on the x-axis.
16:30This point is already on the x-axis.
16:32Its coordinates are 5, 0, and here is
16:34another point already on the x-axis.
16:36Its coordinates are 2, 0, and the
16:39examiner is saying find the length of
16:41AB. Also, for distance in math, this
16:44word is also used: modulus AB. Modulus
16:47AB; the modulus sign is used (we will
16:49study this later in AS Math) because
16:52whatever is inside the modulus, its
16:54result always comes out positive. So,
16:57to show length, distance, or magnitude,
17:00the modulus sign is used in math. So,
17:04if you require the difference, if you
17:06need the distance, common sense says
17:08you should check what directly? Look at
17:10the horizontal gap between these two
17:12points; what is the horizontal gap?
17:14What is the horizontal gap? So, what is
17:15your target? You need the difference
17:17between point A and point B, what is
17:19the horizontal gap between these two?
17:22So you see 1, 2, 3, so the difference
17:25between them is 3 units. The gap
17:28between them is a gap of 3 units. So if
17:31I have to find out what its length will
17:33be? Then its length will be 3 units. So
17:36we should not apply the distance
17:38formula for this. Even if you apply the
17:41distance formula, suppose a student is
17:43weak and they apply the distance
17:45formula, the result will still
17:46ultimately be the same. What are these
17:48points? x1 y1 and what is this point?
17:52x2 y2 so if your target is to apply the
17:54distance formula, then the distance
17:57formula will be 5-2 squared and + 0-0
17:59squared, so what result will be derived
18:02? The result derived will be 9 + 0,
18:05what will be the square root of 9 + 0?
18:08The square root of 9 + 0 will be the
18:10square root of 9, which is 3, by the
18:12way, so the result derived will be 3
18:14units. So this method should not be
18:17applied in this condition. It is common
18:20sense that if you want the horizontal
18:22gap between two points, you should just
18:24count it directly, what is the
18:25difference in their magnitude, clear,
18:27kid? So these two points are already on
18:30the x-axis. If the same situation were
18:33in the exam such that suppose you are
18:36given one point, this point is already,
18:39suppose, 5, 0; this point is already 5,
18:410; this point is on the x-axis and you
18:44have this point, suppose, this is
18:47another point. The coordinates of this
18:49point are, suppose, -2, 0; the
18:52coordinates of this point are, suppose,
18:54-2, 0; now what is your target? Your
18:57target is that you need the difference
18:58between these two points. So now tell
19:00me, what is this gap? What is this gap?
19:03And what is this gap? If I add these
19:05two distances, you will find the total
19:08length. Clear, kid? Total length. So
19:10look, what is this distance you have?
19:12This distance is already 2 units. What
19:15is this distance already? This distance
19:17is 5 units. Because length is always
19:20positive. So if I tell you that, hey, I
19:23want the distance between BA. I want
19:26the distance between BA or the length
19:28of line segment BA. So for BA, we can
19:32directly add them, that we take both of
19:34the horizontal gaps, this difference
19:35and this difference, and what do we do
19:37with both? Add them. So, without the
19:41distance formula, we will know what the
19:44horizontal gap between these two is.
19:47That horizontal gap is 7 units. So, to
19:51summarize, whenever you have points on
19:54the x-axis or on the y-axis, you should
19:57apply basic math to find the vertical
20:00or horizontal difference or distance
20:03between them. So, technically, we
20:06should not apply the distance formula
20:09for this. Is that clear, son? We should
20:11not apply the distance formula because
20:13this is very basic math. Now, let's
20:16continue with the distance formula a
20:17bit further. Look, son, now I am
20:20explaining a new concept. This concept
20:22is called ratio. Ratio. Suppose you
20:26have a line. A point is already given
20:28on it. The coordinates of this point
20:30are A. This point is on your line. The
20:31name of this point is A. Another point
20:33is given here. The name of this point
20:35is X. A, X, and B are your three points
20:38. Now I am explaining a new concept.
20:40The name of this new concept is ratio.
20:43Okay? We must have studied this before
20:46in O-Level Math. The name of this
20:48concept is ratio. Okay? Ratio. We must
20:51have studied this in O-Level Math
20:53itself. Now, look at what a ratio shows
20:55. Suppose the examiner is telling me to
20:58find the ratio. Ratio of lengths. Ratio
21:02of lengths. My target is to find the
21:06ratio of AX and XB. Okay? My target is
21:09to find the ratio of AX and XB. Just
21:12look at the steps, son. This is my
21:15target. Okay? Just as given before.
21:18What is the statement in this question?
21:20The statement in this question is "find
21:22the ratio of AX and XB", that is your
21:25target. Now, notice one thing. All
21:27three of these points are on the same
21:29line. Later, we will study that when
21:32points are on the same line, they are
21:34also called collinear points in
21:35mathematical language. Points on the
21:38same line are also called collinear
21:40points. Now, look at option number one.
21:43What is our target for option number
21:46one? What is our target for option
21:47number one? That we have to find. We
21:50have to find the ratio of AX and XB,
21:52that is our target. The ratio of AX and
21:55XB is our target. We have to find it.
21:58Now, look, one option is this. What do
22:01some weak students do? First, I will
22:03explain that method, that our target is
22:05the ratio. Ratio of what? By the way,
22:08the ratio is of the length. By the way,
22:11in coordinate geometry, the ratio is
22:13always of lengths. Look, we need the
22:15ratio of lengths. The ratio of what? Of
22:18ax and xb. So, what would a weak
22:21student do? And then, what is the best
22:24option? I will explain both options.
22:26Okay? Our target is to get the required
22:29ratio of these two points, ax and xb.
22:31Meaning, what is the ratio between this
22:34length and this length? It is clearly
22:36evident that the length of ax is more
22:38and the length of xb is less. So if I
22:40apply the distance formula to ax, this
22:42is x1. This is y1. Let this be x2 and
22:45this be y2. So if I apply the distance
22:48formula, then according to the formula,
22:51it becomes 11-2 squared and 8-14
22:54squared. Clear, son? I applied the
22:56distance formula. Let's make this a bit
22:58easier. 11-2 means 9 squared and 8-14
23:02means -6 squared. If I simplify this
23:06further, it becomes 81 and 36; if I
23:09simplify this, what is the derived
23:12answer? The derived answer is 17. This
23:17has been found. This length has been
23:19found. The length of a line segment,
23:22whose length has been found? You have
23:24found the length of ax. The length of
23:27ax has been found. Okay, son? The
23:29length of ax has been found. Similarly,
23:31if I find the length of this, xb.
23:35Suppose this point is your x1 y1 and
23:37this point is your x2 y2; your target
23:39is to find the length of XB. So now if
23:42I apply the distance formula in the
23:44same way, how will the distance formula
23:46be applied? I applied the distance
23:48formula just like that. Okay? I am
23:51doing exactly the same thing as we did
23:53before. Okay, son? I am applying the
23:56distance formula. What is the distance
23:59formula? The distance formula you have
24:02is 14-11 x2-x1 squared and y2-y1
24:05squared, this is the distance formula.
24:08If I make this easier, what will 14-11
24:12become? 14-11 will be. See its
24:15difference, 3 squared, which is 9. Let
24:18me show this in complete detail. Let me
24:21show in complete detail what its answer
24:22will be. 3 squared and this will be -2
24:25squared. So if I simplify this further,
24:29it will be 9 + 4, 9 + 4, so I know that
24:33the answer to 9 + 4 is? The answer to 9
24:36+ 4 is 13, the answer is not yet
24:38derived. By the way, the answer is
24:42derived in surd form, surd form. So if
24:45your basic math is strong, that means
24:47what is 13*9? 13*9 = 117. And what is
24:51this? This is root 13. We also studied
24:53this in O-Level Math. Now 13 is a surd.
24:56What will the root of 9 be? The root of
24:599 will be 3. And there is a root 13
25:01surd here as well. So when these two
25:03cancel out, what will be the ratio
25:05derived here? The ratio answer will be
25:083: 1. This is your finally found out
25:11ratio. Meaning, length-wise, this
25:14length is three times this length. So I
25:17can also say technically that the
25:19length of AX is 3 times the length of
25:22XB. Technically, this is what it means.
25:25Okay, now I will show you the best
25:27option for this. I will explain the
25:29best option for this same question. The
25:31option one we used was a very slow
25:34method. Okay, son? The option one I
25:37told you is a weak method. It's a weak
25:40method. So I will explain the best
25:42option for this same question. What is
25:44our target? Our target is the ratio of
25:47AX and XB. Now I am telling you
25:50something new. The ratio of AX and XB,
25:52so look, when these points are
25:54collinear, collinear means points on
25:56the same line. They are collinear
25:58points. We can see that these points
26:00are collinear. We will look at the
26:01meaning of the word collinear in more
26:03detail later. Collinear points means
26:05points on the same line. So I am
26:07telling you a new thing for the ratio.
26:10So the ratio is called the difference
26:12of x-coordinates. Ratio is the
26:15difference of x-coordinates. So now I
26:19will explain to you the best option to
26:21find this ratio. Now you see what the
26:24difference is between the x-coordinates
26:26of these two. Look at this. Come, let
26:28me show you in triangular form what the
26:30difference is between their
26:32x-coordinates. Look, you have a point
26:34like this. This is AX here. And you
26:37have another segment, XB. Okay? There
26:40is another segment, XB. What was your
26:42target? What was required between these
26:45two? You need the distance between
26:47these two. What do you need between
26:49these two? You need the distance. Come,
26:52let me mark this again for a minute. So
26:54what is basically being formed here?
26:57Basically, you have this entire line
26:59segment. What was this whole thing you
27:01had? This was an entire line segment.
27:04Okay, there were three points in it.
27:05One point was named A. One point was
27:08named X and one point was named B. Is
27:10it clear, kid? You had these three
27:12points. A, X, and B. Now tell me, what
27:14is the difference in their
27:16x-coordinates? The x-difference between
27:192 and 11 is nine. And the x-difference
27:23between 11 and 14 is three. So if you
27:27notice, the x-difference between them,
27:30the x-difference between these two is
27:3211-2, which is 9. And the x-difference
27:35between these two, what is the
27:37x-difference between these two? 14-11,
27:41that is 3. So now if I tell you that I
27:45need the ratio, then what is the ratio?
27:48The ratio is the difference of
27:49x-coordinates. Difference of
27:51x-coordinates. So the difference
27:54between A and X is nine. And the
27:56difference between X and B is three. So
27:58when I simplify this, what ratio will I
28:00find? The ratio will be found as 3: 1.
28:04So this is a quick method to find the
28:06ratio. Is it clear, kid? Look again,
28:09what is the gap between 2 and 11? The
28:12gap between the x-coordinates of 2 and
28:1411 is nine, and the gap between the
28:16x-coordinates of 11 and 14 is three.
28:19The gap is three. Is it clear, kid? I
28:21mean, technically, if you notice, this
28:24is two. This is 11. This is 14. Clear?
28:27So now, the gap between two and 11 is
28:30nine, and the gap between 11 and 14 is
28:32three. So what will be the ratio of 9:
28:343? 3: 1. Well. If the points are
28:37collinear, then there is another
28:39similar option for this too. You can do
28:40this exact same thing this way as well.
28:43You can do this same work this way,
28:45that if you want the ratio, you can
28:48also say that the ratio is basically
28:50the difference of y-coordinates. Ratio
28:54is basically the difference of
28:55y-coordinates. Now you see, what ratio
28:58do you want? Your target is the ratio
29:00of AX and XB. Ratio of AX and XB. Now
29:03tell me, what is the gap between their
29:05y-coordinates? 14 and 8, the gap in
29:08their y-coordinate is six. Meaning, the
29:11vertical difference between them is six
29:14. Check it again. Look at the vertical
29:18gap between these two, look at this
29:21vertical gap. This vertical gap between
29:24them is six. This vertical gap is six.
29:27Between what? This point here is your
29:3014. This point here is your eight. So
29:32it's common sense, what will be the gap
29:34between 14 and 8? There will be a gap
29:36of six. Similarly, if you notice, this
29:38y-coordinate is eight and this
29:40y-coordinate is six. What is the
29:42vertical gap between these two? The
29:44vertical gap between these two is two.
29:47Meaning, this vertical gap is smaller.
29:49This vertical gap is two. So, your
29:52target is the ratio of these two. Now,
29:55tell me, what is the vertical
29:56difference between these two? You have
29:59the vertical difference. You have the
30:02vertical difference of 9, and sorry,
30:04you have the vertical difference. This
30:07vertical difference came to six, and
30:09this vertical difference came to two.
30:11So, when I simplify this as well, what
30:14ratio will finally be found? The ratio
30:16will be found to be exactly the same.
30:18So, what is a ratio, basically? This is
30:20a slow method to find the ratio. So,
30:23what is the quick method for the ratio?
30:25Ratio is basically the difference of
30:27x-coordinates or the ratio is basically
30:29the difference of y-coordinates. Okay?
30:32What kind of points were these three
30:33points? These three points were
30:35collinear points. These three points
30:36were collinear points. So, one option
30:38for the ratio is the difference of
30:40x-coordinates. The other option is the
30:42difference of y-coordinates. Look again
30:43at how when you talk about the
30:44difference of x-coordinates. Its
30:46x-coordinate is two. Its x-coordinate
30:48is eleven. Its x-coordinate is fourteen
30:50. So, the gap between two and eleven is
30:52nine. And the gap between eleven and
30:54fourteen is three. So the ratio is 3: 1
30:56. And if you talk about vertical
31:00difference, the vertical difference,
31:02its vertical coordinate is six, its
31:04vertical is eight, its vertical is
31:06fourteen, so the vertical gap between
31:08six and eight is two, and the vertical
31:10gap between eight and fourteen is six.
31:13So vertical gap means the difference of
31:15y-coordinates. So exactly the same
31:17answer is derived, meaning technically,
31:20there is no effect on the distance.
31:22Clear, kid? So both these options are
31:24correct, but obviously the best option
31:26is the one that is the quick option.
31:28Clear, kid? Which is the quick option.
31:31Now, if I give you another example of
31:33this type, of this type, I am just
31:35giving a random example that suppose
31:37you have two or three points like this.
31:39Okay, I am just giving a random example
31:41. Okay, suppose you have this point,
31:45its coordinates are -2, 3 -2, 3, I'm
31:47giving a random example, and suppose
31:50you have this point and its coordinates
31:53are, say, 4, 9, a random example, and
31:55suppose you have this point, its
31:57coordinates are, say, 10, 15, I've just
32:00shown you two or three random points
32:03like this. This point is your A. This
32:06point is your B. This point is your C.
32:09And the examiner is asking you for the
32:11ratio of AB and BC. Your target is the
32:14ratio of AB and BC. So, look at the gap
32:16between the X-coordinates of A and B.
32:19From -2 to 4, the horizontal gap is six
32:22. The horizontal gap is six. There you
32:25go, this gap is a horizontal gap of six
32:27, and what is the horizontal gap from 4
32:29to 10? That gap is also six. That gap
32:31is also six. So from this, I found out
32:34what the ratio technically is. 1: 1,
32:36which means both these lengths are
32:37technically what? The same, because
32:39look, the gap from -2 to 4 is six, and
32:42the gap from 4 to 10 is also six.
32:44Similarly, if you try this question
32:46this way. You know, if you look at
32:48their vertical gap, what is the
32:50vertical gap between 3 and 9? 6, what
32:53is the vertical gap between 9 and 15?
32:57Six. So that means the vertical gap
32:59will also be what? The same. If you
33:02need the ratio of AB and BC, you look
33:04at the vertical difference. What was
33:06the vertical difference derived? 6: 6,
33:08so what is the answer derived? 1: 1
33:12Clear, kid? So, whenever you have
33:15collinear points, points on the same
33:17line, instead of wasting time applying
33:19the distance formula, what is the best
33:21option? The best option is that we
33:24directly do what? We directly look at
33:27the difference of X-coordinates or the
33:29difference of Y-coordinates. Clear, kid
33:33? In the next class, Insha'Allah, we
33:34will continue further. Okay? Allah
33:36Hafiz.

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