Coordinate Geometry

Hello students, welcome to today's mathematics lesson. I'm so happy to be here with you to explore Chapter 7 of your NCERT textbook, which is all about Coordinate Geometry. Now, before we dive into the new concepts, let me remind you of something you learned in Class IX. Do you remember how we locate the position of a point on a plane? Yes, that's right - we need a pair of coordinate axes. The horizontal axis is called the x-axis and the vertical axis is called the y-axis. The distance of a point from the y-axis is called its x-coordinate, or what we call the abscissa. And the distance of a point from the x-axis is called its y-coordinate, or the ordinate. Now, here's an important point to remember - the coordinates of a point on the x-axis are always of the form (x, 0), because its distance from the x-axis is zero. Similarly, the coordinates of a point on the y-axis are always of the form (0, y), because its distance from the y-axis is zero. So students, keep this in mind - whenever you see a point with coordinates (x, 0), it must lie on the x-axis, and whenever you see a point with coordinates (0, y), it must lie on the y-axis.

Now, let me tell you something fascinating about coordinate geometry. In Class IX, you also learned that a linear equation in two variables of the form ax + by + c = 0, where a and b are not both zero, when represented graphically gives a straight line. And in Chapter 2 of your current textbook, you saw that the graph of y = ax² + bx + c, where a is not zero, is a parabola. Isn't it wonderful how algebra and geometry come together in coordinate geometry? It helps us to study geometry using algebra, and at the same time, understand algebra with the help of geometry. Because of this beautiful connection, coordinate geometry is widely used in many fields such as physics, engineering, navigation, seismology, and even art! So what we're learning today is not just some abstract concept - it has real-world applications.

In this chapter, students, you will learn two very important things. First, how to find the distance between two points when their coordinates are given. Second, how to find the coordinates of a point that divides a line segment joining two given points in a given ratio. These are powerful tools that will help you solve many problems. So let's begin with the first concept - the distance formula.

Imagine this situation. There is a town A, and another town B is located 36 kilometers east and 15 kilometers north of town A. Now, how would you find the distance from town A to town B without actually measuring it? This is a practical problem, and we can solve it using coordinate geometry. Let me explain how.

We can represent this situation graphically. Town A can be taken as the origin, which is the point (0, 0). Since town B is 36 kilometers east, that means it is 36 units along the positive x-direction. And since it is 15 kilometers north, it is 15 units along the positive y-direction. So town B has coordinates (36, 15). Now, if we draw a line from A to B, and also draw perpendiculars from A and B to the x-axis, we get a right triangle. The horizontal side of this triangle has length 36 units, and the vertical side has length 15 units. Now, by the Pythagoras theorem, which I hope you remember from your earlier studies, the distance AB squared equals 36 squared plus 15 squared. So AB squared equals 1296 plus 225, which is 1521. And the square root of 1521 is 39. So the distance between town A and town B is 39 kilometers. Wasn't that simple? We didn't need to physically measure the distance - we just used coordinates and the Pythagoras theorem.

Now students, let's build on this idea step by step. First, let's consider the simplest case - when both points lie on the x-axis. Suppose we have two points A(4, 0) and B(6, 0) on the x-axis. Since both points are on the x-axis, their y-coordinates are zero. The distance of point A from the origin is 4 units, and the distance of point B from the origin is 6 units. So the distance from A to B is simply 6 minus 4, which equals 2 units. That makes sense, doesn't it? If two points lie on the x-axis, we can just subtract their x-coordinates to find the distance, taking the absolute value to ensure it's positive.

Similarly, if we take two points on the y-axis, say C(0, 3) and D(0, 8), then the distance between them is 8 minus 3, which equals 5 units. The y-coordinate tells us how far each point is from the origin along the y-axis.

Now, what if the points are not on the same axis? Let's consider point A at (4, 0) on the x-axis and point C at (0, 3) on the y-axis. How do we find the distance from A to C? This is more interesting. We can use the Pythagoras theorem here too. If we draw perpendiculars from A to the y-axis and from C to the x-axis, we get a right triangle with sides of lengths 3 and 4. The distance AC squared equals 3 squared plus 4 squared, which is 9 plus 16, equals 25. So AC equals the square root of 25, which is 5 units. This is a classic 3-4-5 right triangle, students - you might have seen this in your geometry lessons.

Now, let's move to a slightly more general case. Consider two points P(4, 6) and Q(6, 8) that lie in the first quadrant. How do we find the distance between them? Let me guide you through this carefully. First, draw perpendiculars from P and Q to the x-axis. Let these perpendiculars meet the x-axis at points R and S respectively. So R has coordinates (4, 0) and S has coordinates (6, 0). Now, draw a perpendicular from P to the line QS, and let it meet QS at point T. Now, let's identify the key lengths. The horizontal distance RS is 6 minus 4, which equals 2 units. The vertical distance QS is 8 units, and the vertical distance PR is 6 units. So the vertical distance QT is 8 minus 6, which equals 2 units. And PT is equal to RS, which is 2 units. Now, in the right triangle PTQ, we have PT equal to 2 and QT equal to 2. By the Pythagoras theorem, PQ squared equals PT squared plus QT squared, which is 2 squared plus 2 squared, equals 4 plus 4, equals 8. So PQ equals the square root of 8, which is 2 times the square root of 2, or approximately 2.83 units.

Now students, what if the two points are in different quadrants? Let's consider points P(6, 4) and Q(-5, -3). Here, P is in the first quadrant and Q is in the third quadrant. To find the distance between them, we follow a similar process. Draw a perpendicular from Q to the x-axis, and let it meet the x-axis at S. Now, draw a perpendicular from P to the line QS, and let it meet the y-axis at R. Now, let's find the lengths. The horizontal distance PT is the difference between the x-coordinates, which is 6 minus (-5), equals 11 units. The vertical distance QT is the difference between the y-coordinates, which is 4 minus (-3), equals 7 units. Now, in the right triangle PTQ, by the Pythagoras theorem, PQ squared equals 11 squared plus 7 squared, which is 121 plus 49, equals 170. So PQ equals the square root of 170, which is approximately 13.04 units.

Now we are ready to derive the general distance formula. Let P with coordinates (x₁, y₁) and Q with coordinates (x₂, y₂) be any two points on the coordinate plane. We want to find the distance between them. Draw perpendiculars from P and Q to the x-axis, meeting it at points R and S respectively. Now, draw a perpendicular from P to the line QS, and let it meet QS at point T. Now, let's find the key lengths. The horizontal distance RS is the difference between the x-coordinates, which is x₂ minus x₁. But this is also equal to PT. The vertical distance QT is the difference between the y-coordinates, which is y₂ minus y₁. Now, in the right triangle PTQ, by the Pythagoras theorem, we have PQ squared equals PT squared plus QT squared. Substituting our values, we get PQ squared equals (x₂ minus x₁) squared plus (y₂ minus y₁) squared. Therefore, PQ equals the square root of ((x₂ minus x₁) squared plus (y₂ minus y₁) squared). This is called the distance formula, students. The distance between any two points P(x₁, y₁) and Q(x₂, y₂) is given by this formula.

Now, a very important point to remember - since distance is always non-negative, we take only the positive square root. Also, note that we can write the formula in another way: PQ equals the square root of ((x₁ minus x₂) squared plus (y₁ minus y₂) squared). This is because squaring eliminates the sign, so (x₂ minus x₁) squared is the same as (x₁ minus x₂) squared. Both forms are correct.

Let me give you a special case. What is the distance of a point P(x, y) from the origin O(0, 0)? Well, using our formula with (x₁, y₁) as (0, 0) and (x₂, y₂) as (x, y), we get OP equals the square root of (x squared plus y squared). So the distance from the origin is simply the square root of the sum of the squares of the coordinates. This is a very useful result, students - keep it in mind.

Now, let's work through some examples to solidify our understanding. In Example 1, we need to check whether the points (3, 2), (-2, -3), and (2, 3) form a triangle. If so, we need to name the type of triangle. Let me call these points P(3, 2), Q(-2, -3), and R(2, 3). First, let's find the distances between each pair of points. Using the distance formula, PQ equals the square root of ((3 minus (-2)) squared plus (2 minus (-3)) squared). That's the square root of (5 squared plus 5 squared), which is the square root of (25 plus 25), equals the square root of 50, which is approximately 7.07 units. Next, QR equals the square root of ((-2 minus 2) squared plus (-3 minus 3) squared). That's the square root of ((-4) squared plus (-6) squared), which is the square root of (16 plus 36), equals the square root of 52, approximately 7.21 units. And PR equals the square root of ((3 minus 2) squared plus (2 minus 3) squared), which is the square root of (1 squared plus (-1) squared), equals the square root of (1 plus 1), equals the square root of 2, approximately 1.41 units. Now, to form a triangle, the sum of any two sides must be greater than the third side. Let's check: 7.07 plus 1.41 equals 8.48, which is greater than 7.21. 7.07 plus 7.21 equals 14.28, which is greater than 1.41. And 7.21 plus 1.41 equals 8.62, which is greater than 7.07. So yes, these three points do form a triangle. Now, to determine the type of triangle, let's check if it satisfies the Pythagoras theorem. We have PQ squared plus PR squared equals 50 plus 2 equals 52, and QR squared equals 52. So PQ squared plus PR squared equals QR squared. By the converse of the Pythagoras theorem, this means that the angle at P is 90 degrees. So triangle PQR is a right triangle. That's Example 1, students.

Now, let's look at Example 2. We need to show that the points (1, 7), (4, 2), (-1, -1), and (-4, 4) are the vertices of a square. Let me call these points A(1, 7), B(4, 2), C(-1, -1), and D(-4, 4). One way to show that ABCD is a square is to use the property that all four sides are equal and both diagonals are also equal. Let's find the distances. AB equals the square root of ((1 minus 4) squared plus (7 minus 2) squared), which is the square root of ((-3) squared plus 5 squared), equals the square root of (9 plus 25), equals the square root of 34. BC equals the square root of ((4 minus (-1)) squared plus (2 minus (-1)) squared), which is the square root of (5 squared plus 3 squared), equals the square root of (25 plus 9), equals the square root of 34. CD equals the square root of ((-1 minus (-4)) squared plus (-1 minus 4) squared), which is the square root of (3 squared plus (-5) squared), equals the square root of (9 plus 25), equals the square root of 34. And DA equals the square root of ((1 minus (-4)) squared plus (7 minus 4) squared), which is the square root of (5 squared plus 3 squared), equals the square root of (25 plus 9), equals the square root of 34. So all four sides are equal, each being the square root of 34 units. Now let's find the diagonals. AC equals the square root of ((1 minus (-1)) squared plus (7 minus (-1)) squared), which is the square root of (2 squared plus 8 squared), equals the square root of (4 plus 64), equals the square root of 68. And BD equals the square root of ((4 minus (-4)) squared plus (2 minus 4) squared), which is the square root of (8 squared plus (-2) squared), equals the square root of (64 plus 4), equals the square root of 68. So both diagonals are equal, each being the square root of 68 units. Since all four sides are equal and both diagonals are equal, the quadrilateral ABCD is a square. That's one way to solve it.

There's also an alternative solution. We can find the four sides and one diagonal, say AC. We already have all four sides equal to the square root of 34. Now, AD squared plus DC squared equals 34 plus 34 equals 68, and AC squared equals 68. So AD squared plus DC squared equals AC squared. By the converse of the Pythagoras theorem, angle D is 90 degrees. A quadrilateral with all four sides equal and one angle 90 degrees is a square. So ABCD is indeed a square. Both methods lead to the same conclusion, students.

Now, let's look at Example 3. This is a practical problem about seating in a classroom. The points A(3, 1), B(6, 4), and C(8, 6) represent the seats of Ashima, Bharti, and Camella respectively. We need to determine if they are seated in a straight line. Let's find the distances between each pair of points. AB equals the square root of ((6 minus 3) squared plus (4 minus 1) squared), which is the square root of (3 squared plus 3 squared), equals the square root of (9 plus 9), equals the square root of 18, which is 3 times the square root of 2. BC equals the square root of ((8 minus 6) squared plus (6 minus 4) squared), which is the square root of (2 squared plus 2 squared), equals the square root of (4 plus 4), equals the square root of 8, which is 2 times the square root of 2. And AC equals the square root of ((8 minus 3) squared plus (6 minus 1) squared), which is the square root of (5 squared plus 5 squared), equals the square root of (25 plus 25), equals the square root of 50, which is 5 times the square root of 2. Now, let's check if AB plus BC equals AC. AB plus BC equals 3 times the square root of 2 plus 2 times the square root of 2, which equals 5 times the square root of 2, and that equals AC. So yes, the points A, B, and C are collinear. Therefore, they are seated in a straight line. This makes sense, doesn't it? If three points are collinear, then the distance from the first to the third equals the sum of the distances from the first to the second and from the second to the third.

Now, let's look at Example 4. We need to find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5). Let me call the point P(x, y). We are given that the distance from P to A(7, 1) equals the distance from P to B(3, 5). In other words, AP equals BP. Squaring both sides, we get AP squared equals BP squared. Now, using the distance formula, AP squared equals (x minus 7) squared plus (y minus 1) squared, and BP squared equals (x minus 3) squared plus (y minus 5) squared. Setting these equal, we get (x minus 7) squared plus (y minus 1) squared equals (x minus 3) squared plus (y minus 5) squared. Expanding both sides, we get x squared minus 14x plus 49 plus y squared minus 2y plus 1 equals x squared minus 6x plus 9 plus y squared minus 10y plus 25. Simplifying, we get x squared minus 14x plus y squared minus 2y plus 50 equals x squared minus 6x plus y squared minus 10y plus 34. Subtracting x squared and y squared from both sides, we get minus 14x minus 2y plus 50 equals minus 6x minus 10y plus 34. Bringing all terms to one side, we get minus 14x plus 6x minus 2y plus 10y plus 50 minus 34 equals 0. Simplifying, we get minus 8x plus 8y plus 16 equals 0. Dividing by 8, we get minus x plus y plus 2 equals 0, or x minus y equals 2. So the required relation is x minus y equals 2. Now, here's an interesting observation. You learned in your earlier studies that a point which is equidistant from two points A and B lies on the perpendicular bisector of the line segment AB. So the graph of the equation x minus y equals 2 is actually the perpendicular bisector of the line segment joining A(7, 1) and B(3, 5). This is a very useful geometric interpretation, students.

Now, let's look at Example 5. We need to find a point on the y-axis which is equidistant from the points A(6, 5) and B(-4, 3). Now, a point on the y-axis has an x-coordinate of 0. So let the point be P(0, y). We need AP equals BP. Using the distance formula, AP squared equals (6 minus 0) squared plus (5 minus y) squared, which is 36 plus (5 minus y) squared. BP squared equals (-4 minus 0) squared plus (3 minus y) squared, which is 16 plus (3 minus y) squared. Setting these equal, we get 36 plus (5 minus y) squared equals 16 plus (3 minus y) squared. Expanding, we get 36 plus 25 minus 10y plus y squared equals 16 plus 9 minus 6y plus y squared. Simplifying, we get 61 minus 10y plus y squared equals 25 minus 6y plus y squared. Subtracting y squared from both sides, we get 61 minus 10y equals 25 minus 6y. So 61 minus 25 equals 10y minus 6y, which gives 36 equals 4y, so y equals 9. Therefore, the required point is (0, 9). Let's verify our answer. AP equals the square root of ((6 minus 0) squared plus (5 minus 9) squared), which is the square root of (36 plus (-4) squared), equals the square root of (36 plus 16), equals the square root of 52. And BP equals the square root of ((-4 minus 0) squared plus (3 minus 9) squared), which is the square root of (16 plus (-6) squared), equals the square root of (16 plus 36), equals the square root of 52. So yes, both distances are equal. And as we noted in the previous example, this point (0, 9) is the intersection of the y-axis and the perpendicular bisector of AB.

Now students, let's move on to the next major concept in this chapter, which is the Section Formula. This is about finding the coordinates of a point that divides a line segment joining two given points in a given ratio.

Let me start with a practical situation. Suppose a telephone company wants to position a relay tower at point P between towns A and B in such a way that the distance of the tower from B is twice its distance from A. In other words, if P lies on the line segment AB, it will divide AB in the ratio 1:2, where AP:PB equals 1:2. Now, if we take A as the origin O, and 1 kilometer as one unit on both axes, then the coordinates of B will be (36, 15), as we discussed earlier. Now, to know the position of the tower, we must find the coordinates of P. How do we do that? Let me show you.

Let the coordinates of P be (x, y). Draw perpendiculars from P and B to the x-axis, meeting it at points D and E respectively. Draw a perpendicular from P to the line BE, meeting it at point C. Now, by the AA similarity criterion, which you studied in Chapter 6, triangle POD is similar to triangle BPC. Therefore, the ratio of corresponding sides is equal. Specifically, OD divided by PC equals OP divided by PB, which is 1/2. Also, PD divided by BC equals OP divided by PB, which is 1/2. From the first ratio, OD/PC equals 1/2, which means x/(36 minus x) equals 1/2. From the second ratio, PD/BC equals 1/2, which means y/(15 minus y) equals 1/2. Solving these equations, we get 2x equals 36 minus x, so 3x equals 36, so x equals 12. And 2y equals 15 minus y, so 3y equals 15, so y equals 5. So the coordinates of P are (12, 5). Let's verify: the distance from A to P is the square root of (12 squared plus 5 squared), which is the square root of (144 plus 25), equals the square root of 169, which is 13. And the distance from P to B is the square root of ((36 minus 12) squared plus (15 minus 5) squared), which is the square root of (24 squared plus 10 squared), equals the square root of (576 plus 100), equals the square root of 676, which is 26. So indeed, AP:PB equals 13:26, which is 1:2. So our answer is correct.

Now, let's generalize this. Consider any two points A(x₁, y₁) and B(x₂, y₂), and assume that P(x, y) divides AB internally in the ratio m₁:m₂, that is, PA/PB equals m₁/m₂. We want to find the coordinates of P. Draw perpendiculars from A, P, and B to the x-axis, meeting it at points R, S, and T respectively. Draw a line from A parallel to the x-axis, meeting the line through P at point Q. Now, by the AA similarity criterion, triangle PAQ is similar to triangle BPC. Therefore, PA/BP equals AQ/PC equals PQ/BC. Now, let's find these lengths. AQ equals RS, which is OS minus OR, so that's x minus x₁. PC equals ST, which is OT minus OS, so that's x₂ minus x. PQ equals PS minus QS, which is y minus y₁. And BC equals BT minus CT, which is y₂ minus y. Substituting these values in the ratio, we get m₁/m₂ equals (x minus x₁)/(x₂ minus x) equals (y minus y₁)/(y₂ minus y). Taking the first part, m₁/m₂ equals (x minus x₁)/(x₂ minus x). Cross-multiplying, we get m₁(x₂ minus x) equals m₂(x minus x₁). Expanding, we get m₁x₂ minus m₁x equals m₂x minus m₂x₁. Bringing the x terms to one side, we get m₁x₂ plus m₂x₁ equals m₁x plus m₂x, which is (m₁ plus m₂)x. So x equals (m₁x₂ plus m₂x₁) divided by (m₁ plus m₂). Similarly, from the second part, we get y equals (m₁y₂ plus m₂y₁) divided by (m₁ plus m₂). So the coordinates of the point P(x, y) which divides the line segment joining the points A(x₁, y₁) and B(x₂, y₂), internally, in the ratio m₁:m₂ are ((m₁x₂ plus m₂x₁)/(m₁ plus m₂), (m₁y₂ plus m₂y₁)/(m₁ plus m₂)). This is known as the section formula, students. This is a very important formula, so please remember it.

Now, if the ratio in which P divides AB is k:1, then the coordinates of P will be ((kx₂ plus x₁)/(k plus 1), (ky₂ plus y₁)/(k plus 1)). This is just a special case of the section formula where m₁ equals k and m₂ equals 1.

Now, here's a special case. What is the midpoint of a line segment? The midpoint divides the line segment in the ratio 1:1. So if we put m₁ equals 1 and m₂ equals 1 in the section formula, we get x equals (1 times x₂ plus 1 times x₁) divided by (1 plus 1), which is (x₁ plus x₂)/2. Similarly, y equals (y₁ plus y₂)/2. So the coordinates of the midpoint of the line segment joining the points A(x₁, y₁) and B(x₂, y₂) are ((x₁ plus x₂)/2, (y₁ plus y₂)/2). This is called the midpoint formula. It's very useful for finding the center of a line segment, and we'll see its applications in some examples.

Now, let's work through some examples to understand the section formula better. In Example 6, we need to find the coordinates of the point which divides the line segment joining the points (4, -3) and (8, 5) in the ratio 3:1 internally. Let P(x, y) be the required point. Using the section formula with m₁ equals 3 and m₂ equals 1, we get x equals (3 times 8 plus 1 times 4) divided by (3 plus 1), which is (24 plus 4)/4, equals 28/4, equals 7. And y equals (3 times 5 plus 1 times (-3)) divided by (3 plus 1), which is (15 minus 3)/4, equals 12/4, equals 3. So the required point is (7, 3). That's straightforward, students.

Now, let's look at Example 7. We need to find the ratio in which the point (-4, 6) divides the line segment joining the points A(-6, 10) and B(3, -8). Let the point (-4, 6) divide AB internally in the ratio m₁:m₂. Using the section formula, we have (-4, 6) equals ((3m₁ minus 6m₂)/(m₁ plus m₂), (-8m₁ plus 10m₂)/(m₁ plus m₂)). Since two points are equal if their corresponding coordinates are equal, we get two equations: -4 equals (3m₁ minus 6m₂)/(m₁ plus m₂), and 6 equals (-8m₁ plus 10m₂)/(m₁ plus m₂). Let's solve the first equation. Multiplying both sides by (m₁ plus m₂), we get -4(m₁ plus m₂) equals 3m₁ minus 6m₂. Expanding, we get -4m₁ minus 4m₂ equals 3m₁ minus 6m₂. Bringing all terms to one side, we get -4m₁ minus 3m₁ plus (-4m₂ plus 6m₂) equals 0, which is -7m₁ plus 2m₂ equals 0, or 7m₁ equals 2m₂. So m₁:m₂ equals 2:7. Let's verify with the second equation. We need to check if the y-coordinate also satisfies this ratio. Using m₁/m₂ equals 2/7, we can write the second equation as (-8m₁/m₂ plus 10)/(m₁/m₂ plus 1). Substituting m₁/m₂ equals 2/7, we get (-8 times 2/7 plus 10) divided by (2/7 plus 1), which is (-16/7 plus 10) divided by (2/7 plus 7/7), equals ((-16 plus 70)/7) divided by (9/7), equals (54/7) divided by (9/7), equals 54/7 times 7/9, equals 54/9, equals 6. Yes, it matches. So the point (-4, 6) divides the line segment joining A(-6, 10) and B(3, -8) in the ratio 2:7.

There's an alternative approach. We can assume the ratio is k:1. Then using the section formula, we get (-4, 6) equals ((3k minus 6)/(k plus 1), (-8k plus 10)/(k plus 1)). From the x-coordinate, -4 equals (3k minus 6)/(k plus 1). Cross-multiplying, we get -4(k plus 1) equals 3k minus 6, so -4k minus 4 equals 3k minus 6, so -4k minus 3k equals -6 plus 4, so -7k equals -2, so k equals 2/7. So the ratio is (2/7):1, which is 2:7. Same answer!

Now, let's look at Example 8. We need to find the coordinates of the points of trisection of the line segment joining the points A(2, -2) and B(-7, 4). Points of trisection are points that divide the line segment into three equal parts. So let P and Q be the points of trisection, such that AP equals PQ equals QB. This means P divides AB in the ratio 1:2, and Q divides AB in the ratio 2:1. Let's find P first. Using the section formula with m₁ equals 1 and m₂ equals 2, we get the coordinates of P as ((1 times (-7) plus 2 times 2)/(1 plus 2), (1 times 4 plus 2 times (-2))/(1 plus 2)), which is ((-7 plus 4)/3, (4 minus 4)/3), equals (-3/3, 0/3), equals (-1, 0). So P is (-1, 0). Now, for Q, using m₁ equals 2 and m₂ equals 1, we get ((2 times (-7) plus 1 times 2)/(2 plus 1), (2 times 4 plus 1 times (-2))/(2 plus 1)), which is ((-14 plus 2)/3, (8 minus 2)/3), equals (-12/3, 6/3), equals (-4, 2). So Q is (-4, 2). So the points of trisection are (-1, 0) and (-4, 2). We could also have obtained Q by noting that it is the midpoint of PB, so we could have used the midpoint formula to find it.

Now, let's look at Example 9. We need to find the ratio in which the y-axis divides the line segment joining the points (5, -6) and (-1, -4). Also, we need to find the point of intersection. Let the ratio be k:1. Then by the section formula, the coordinates of the point which divides AB in the ratio k:1 are ((-k plus 5)/(k plus 1), (-4k minus 6)/(k plus 1)). Now, this point lies on the y-axis, and we know that on the y-axis, the x-coordinate, which is the abscissa, is 0. So we have (-k plus 5)/(k plus 1) equals 0. This implies -k plus 5 equals 0, so k equals 5. So the ratio is 5:1. Now, to find the point of intersection, we substitute k equals 5 into the y-coordinate formula. We get y equals (-4 times 5 minus 6)/(5 plus 1), which is (-20 minus 6)/6, equals -26/6, equals -13/3. So the point of intersection is (0, -13/3). That's Example 9.

Now, let's look at Example 10. We have vertices A(6, 1), B(8, 2), C(9, 4), and D(p, 3) of a parallelogram, taken in order. We need to find the value of p. Now, an important property of a parallelogram is that its diagonals bisect each other. That means the midpoint of AC equals the midpoint of BD. Let's find the midpoint of AC. The coordinates are ((6 plus 9)/2, (1 plus 4)/2), which is (15/2, 5/2). Now, the midpoint of BD is ((8 plus p)/2, (2 plus 3)/2), which is ((8 plus p)/2, 5/2). Setting these equal, we get (15/2, 5/2) equals ((8 plus p)/2, 5/2). So 15/2 equals (8 plus p)/2, which implies 15 equals 8 plus p, so p equals 7. That's the answer.

Now students, we have covered all the main concepts in this chapter. Let me summarize what we have learned today.

In this chapter on Coordinate Geometry, we studied several important concepts. First, we learned the distance formula, which allows us to find the distance between any two points P(x₁, y₁) and Q(x₂, y₂). The formula is PQ equals the square root of ((x₂ minus x₁) squared plus (y₂ minus y₁) squared). We also learned that the distance of a point P(x, y) from the origin O(0, 0) is given by OP equals the square root of (x squared plus y squared). We worked through several examples to understand how to apply the distance formula to find distances between points, to check if three points form a triangle and determine its type, to verify if four points form a square, to check if points are collinear, and to find points that are equidistant from two given points.

Then we moved on to the section formula, which helps us find the coordinates of a point that divides a line segment joining two given points in a given ratio. The coordinates of the point P(x, y) which divides the line segment joining the points A(x₁, y₁) and B(x₂, y₂), internally, in the ratio m₁:m₂ are ((m₁x₂ plus m₂x₁)/(m₁ plus m₂), (m₁y₂ plus m₂y₁)/(m₁ plus m₂)). We also learned the special case of the midpoint formula, which gives the coordinates of the midpoint of a line segment as ((x₁ plus x₂)/2, (y₁ plus y₂)/2). We worked through examples to find points that divide line segments in given ratios, to find points of trisection, to find where the coordinate axes divide line segments, and to use the property that diagonals of a parallelogram bisect each other to find unknown coordinates.

These are powerful tools in coordinate geometry, and you will find them very useful not only in your exams but also in solving many practical problems in the future. Remember, practice is key to mastering these concepts. So make sure you work through the examples and exercises in your textbook.

That's all for today, students. Thank you for your attention, and I hope you enjoyed learning about coordinate geometry as much as I enjoyed teaching it to you. Keep practicing, and goodbye!