Brief Summary
Alright ji, toh yeh hai Coordinate Geometry ka day 13 of the 14-day marathon by Mission JEET. Sir talks about plotting points on the Cartesian plane, distance formula, section formula, and how to apply these concepts to solve problems, including some tricky ones involving quadrilaterals and collinearity. Also, there's a pep talk about not getting bogged down by tough questions and focusing on content over flashy names.
- Basics of Coordinate Geometry
- Distance and Section Formulas
- Application to Various Geometric Problems
Introduction and Schedule
Sir welcomes everyone to Day 13 of the 14-day marathon, focusing on Coordinate Geometry. He apologizes for being late due to traffic and other work commitments. He outlines the schedule: February 13th and 14th will complete the 14-day marathon, followed by Maha-Marathons on the 15th and 16th to brush up the entire Maths syllabus. A final touch-up session is scheduled for February 17th at 4:00 AM, covering all formulas and important questions.
Cartesian Plane and Basic Concepts
Sir explains the basics of coordinate geometry, including the x-axis, y-axis, and the origin on a graph paper. The entire plane is referred to as the Cartesian plane. He explains how to plot a point, say P (2, 3), by moving 2 units on the x-axis and 3 units on the y-axis. Similarly, he plots another point R (-2, -4).
Quadrants and Distance from Axes
The x and y axes divide the plane into four quadrants: first, second, third, and fourth. He explains how to identify the quadrant in which a point lies. The distance of a point P from the x-axis is equal to its y-coordinate, and the distance from the y-axis is equal to its x-coordinate, always taken as positive values.
Abscissa, Ordinate, and Lallu Problem
The x-coordinate is also known as the abscissa, and the y-coordinate is known as the ordinate. Sir introduces a "Lallu Problem" where students have to find the abscissa, ordinate, distance from the y-axis, distance from the x-axis, and the quadrant of a point S (-3, 4) without plotting it on a graph. He then explains the sign conventions for each quadrant.
Equations of Axes and Random Points
Sir explains that on the x-axis, the y-coordinate is always zero, hence the equation of the x-axis is y = 0. Similarly, on the y-axis, the x-coordinate is always zero, so the equation of the y-axis is x = 0. A random point on the x-axis can be represented as (x, 0), and a random point on the y-axis as (0, y).
Distance Formula Introduction
Sir transitions to the distance formula, which is used to calculate the distance between two points P and R. He states the formula: D = √((x2 - x1)² + (y2 - y1)²). He mentions that coordinate geometry mainly revolves around the distance formula and the section formula.
Applying the Distance Formula
Sir demonstrates how to apply the distance formula with an example. Given points P (-4, 7) and Q (2, -5), he calculates the distance between them using the formula. He simplifies the result to 6√5 units. He also clarifies that it doesn't matter whether you use (x2 - x1) or (x1 - x2) because the square will always make the result positive.
Distance from X-axis Quick Question
Sir conducts a quick poll to check students' understanding. The question is to find the distance of a point (3, 7) from the x-axis. The correct answer is 7, as the distance from the x-axis is simply the y-coordinate.
Finding X Value with Distance Given
Sir presents a problem where the distance between points A (0, 0) and B (x, -4) is 5 units. Students need to find the value(s) of x. By applying the distance formula and solving the equation, the values of x are found to be +3 and -3.
Distance from X-axis Conceptual Question
Sir poses a conceptual question: The distance of a point A from the x-axis is 3 units. Which of the following cannot be the coordinates of point A? The options are (3,3), (-3,3), (0,-3), (2,3), (2,-3), (4,1). The correct answer is (4,1), as the y-coordinate must be either 3 or -3.
Circle and Distance Formula Problem
Sir presents a problem involving a circle. The center of the circle is (2a, a-7), and the circle passes through the point (11, -9). The radius of the circle is 5√2. The task is to find the value of a. By using the distance formula to equate the radius and simplifying the equation, a quadratic equation is formed. Solving the quadratic equation gives the values of a as 3 and 5.
Equidistant Points Problem
Sir explains a problem where a point P (k-1, 2) is equidistant from points A (3, k) and B (k, 5). The task is to find the values of k. By equating the distances PA and PB, squaring both sides, and simplifying, a quadratic equation is formed. Solving the quadratic equation gives the values of k as 5 and 1.
Equidistant Points with a Twist
Sir presents a problem where the x-coordinate of a point P is twice its y-coordinate. Point P is equidistant from Q (2, -5) and R (-3, 6). The task is to find the coordinates of P. By setting up the equation PQ = PR, squaring both sides, and simplifying, the y-coordinate is found to be 8. Therefore, the coordinates of P are (16, 8).
Proving a Relationship with Equidistant Points
Sir presents a problem where a point P (x, y) is equidistant from points A (a+b, a-b) and B (a-b, a+b). The task is to prove that bx = ay. By equating the distances PA and PB, squaring both sides, and simplifying, the relationship bx = ay is proven.
Finding a Point on the Y-axis
Sir presents a problem where the task is to find a point on the y-axis that is equidistant from points A (5, -2) and B (-3, 2). By recognizing that any point on the y-axis has coordinates (0, y), the problem is simplified. Equating the distances PA and PB, squaring both sides, and solving for y gives the y-coordinate as -2. Therefore, the point is (0, -2).
Distance Formula Easy Question
Sir gives a very easy question to find the distance between the points (a-b, 0) and (0, a+b). The correct answer is √(2a² + 2b²).
Distance Formula Trigonometry Question
Sir presents a problem to find the distance between the points (acosθ + bsinθ, 0) and (0, asinθ - bcosθ). By applying the distance formula and using the trigonometric identity sin²θ + cos²θ = 1, the distance is found to be √(a² + b²).
Showing Triangles are Similar
Sir presents a problem to show that triangle ABC with vertices A(-2, 0), B(2, 0), and C(0, 2) is similar to triangle PQR with vertices P(-4, 0), Q(4, 0), and R(0, 4). By calculating the lengths of the sides of both triangles and showing that the ratios of corresponding sides are equal, it is proven that the triangles are similar by the Side-Side-Side (SSS) similarity criterion.
Figures and Coordinate Geometry
Sir discusses how coordinate geometry can be used to prove properties of different geometric figures. He provides a summary of how to prove different types of quadrilaterals and triangles:
- Parallelogram: Show that both pairs of opposite sides are equal in length.
- Rectangle: Show that opposite sides are equal in length and the diagonals are equal in length.
- Square: Show that all sides are equal in length and the diagonals are equal in length.
- Rhombus: Show that all sides are equal in length.
- Right Triangle: Show that the Pythagorean theorem holds (a² + b² = c²).
- Isosceles Triangle: Show that any two sides are equal in length.
- Equilateral Triangle: Show that all three sides are equal in length.
Isosceles Right Triangle Problem
Sir presents a problem to prove that the points A(7, 10), B(-2, 5), and C(3, -4) are vertices of an isosceles right triangle. By calculating the lengths of the sides AB, BC, and AC, it is shown that AB = BC (isosceles) and that AC² = AB² + BC² (right triangle).
Right Angled Triangle Problem
Sir presents a problem where P(x, y), Q(-2, -3), and R(2, 3) are vertices of a right triangle, right-angled at P. The task is to find the relationship between x and y. By applying the Pythagorean theorem (RQ² = PR² + PQ²) and simplifying, the relationship x² + y² = 13 is found.
Equilateral Triangle Tricky Problem
Sir presents a challenging problem: Two vertices of an equilateral triangle are (-5, 3) and (5, 3). Find the coordinates of the third vertex, given that the origin lies inside the triangle. This problem requires careful consideration of the geometry and the position of the origin. The third vertex is found to be (0, 3 - 5√3).
Collinear Points Problem
Sir explains how to prove that three points are collinear. The condition for collinearity is that the sum of any two distances between the points must be equal to the third distance.
Section Formula Problem
Sir explains how to apply the section formula with an example. Given points A (2, -2) and B (-7, 4), and a point P that divides the line segment AB in the ratio 1:2, find the coordinates of P.
Section Formula Problem with A and B
Sir explains how to apply the section formula with an example. Given points A (3a+1, -3) and B (8a, 5) and a point P (9a-2, -b) that divides the line segment AB in the ratio 3:1, find the values of a and b.
Four Equal Parts Problem
Sir explains how to find the coordinates of points that divide a line segment into four equal parts. The key is to consider each point separately and combine the ratios on either side.
Ratio Problem
Sir explains how to find the ratio in which a point divides a line segment. The key is to express the ratio as k:1 and solve for k.
Mid Point Formula
Sir explains the mid-point formula. The mid-point of the line segment joining the points (x1, y1) and (x2, y2) is ((x1+x2)/2, (y1+y2)/2).
Mid Point Formula Problem
Sir explains how to apply the mid-point formula with an example. Given points A (-4, 5) and B (4, 6), find the mid-point of the line segment AB.
Parallelogram Problem
Sir explains how to apply the mid-point formula with an example. Given points A (6, 1), B (8, 2), C (9, 4) and D (7, 3) are the vertices of a parallelogram, find the values of p and q.
Line Produced Problem
Sir explains how to apply the mid-point formula with an example. Given points A (-1, 7) and B (4, -3) and a point C which lies on the line AB produced such that AC = 2BC, find the coordinates of C.
Assertion Reason Problem
Sir explains how to solve assertion reason problems. The key is to first check if the assertion is true or false. If the assertion is false, then the answer is D. If the assertion is true, then check if the reason is true and if the reason is a correct explanation of the assertion.
Case Study Problem
Sir explains how to solve case study problems. The key is to read the problem carefully and identify the relevant information. Then, apply the appropriate formulas and concepts to solve the problem.
Final Words
Sir concludes the lecture, emphasizing the importance of practicing all the questions from the PDF. He reminds students about the upcoming Maha-Marathons and encourages them to stay focused and not give up, even when faced with difficult problems.
