{"id":156320,"date":"2022-05-03T17:00:40","date_gmt":"2022-05-03T11:30:40","guid":{"rendered":"https:\/\/www.aakash.ac.in\/blog\/?p=156320"},"modified":"2023-03-27T14:40:15","modified_gmt":"2023-03-27T09:10:15","slug":"class-10-coordinate-geometry-exercise-notes-for-exams","status":"publish","type":"post","link":"https:\/\/www.aakash.ac.in\/blog\/class-10-coordinate-geometry-exercise-notes-for-exams\/","title":{"rendered":"Class 10 Coordinate Geometry Exercise Notes for Exams"},"content":{"rendered":"<p><span style=\"font-weight: 400;\">Class 10 Mathematics Exercise Notes for Chapter 7 &#8211; Coordinate Geometry describes the various core concepts of coordinate geometry. Coordinate Geometry helps the students to interpret and quantitatively present geometrical shapes. It helps to extract numerical information by taking out logical conclusions. We have Class 10 Mathematics Exercise Notes curated by Expert Teachers. These are helpful for students in solving various exercises of this chapter for homework or exam practice. Also, these concepts build a foundation for further study &amp; competitive exams like <\/span><a href=\"https:\/\/www.aakash.ac.in\/national-talent-search-examination-ntse\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">NTSE<\/span><\/a><span style=\"font-weight: 400;\">, <\/span><a href=\"https:\/\/www.aakash.ac.in\/olympiads-gateway-global-recognition\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">Olympiad<\/span><\/a><span style=\"font-weight: 400;\"> exam, <\/span><a href=\"https:\/\/www.aakash.ac.in\/kvpy-exam\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">KVPY<\/span><\/a><span style=\"font-weight: 400;\">, and more. The students can get easy access to our exercise notes for different topics. By practicing with our explanatory notes, students can attain perfection in Coordinate geometry. Also, these notes come in handy for your exam preparation. So, let us begin with it!<\/span><\/p>\n<p>Are you looking for the latest information and how to download JEE admit cards? Check here and download <a href=\"https:\/\/www.aakash.ac.in\/jee-main-admit-card\" target=\"_blank\" rel=\"noopener\">JEE Main 2022 Admit Card<\/a> and <a href=\"https:\/\/www.aakash.ac.in\/cucet-admit-card\" target=\"_blank\" rel=\"noopener\">JEE Advanced 2022 Admit Card.<\/a><\/p>\n<h3>What is Coordinate Geometry?<\/h3>\n<p><span style=\"font-weight: 400;\">The branch of mathematics that deals with cartesian coordinates is known as Coordinate Geometry. Coordinate geometry has evolved as an algebraic means for studying the geometry of figures. It simply helps to study geometry by utilizing algebra and understand algebra by utilizing concepts of geometry. This system is most commonly used to exploit the equations of several two-dimensional figures like circles, triangles, squares, and more.<\/span><\/p>\n<h3>What will Students get to learn?<\/h3>\n<p><span style=\"font-weight: 400;\">Class 10 Coordinate Geometry is a very crucial chapter in mathematics. It will help the students to find:<\/span><\/p>\n<ul>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">The distance between the two points when their coordinates are given,<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">The area of the triangle formed by three given points,<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">The coordinates of the point which divides a line segment that joins the given points in a given ratio.\u00a0<\/span><\/li>\n<\/ul>\n<p><span style=\"font-weight: 400;\">The students must gain a good understanding of this lesson as it forms the base for various other related topics like Constructions in Class 10 Mathematics. And this can be done if you know the detailed <\/span><a href=\"https:\/\/www.aakash.ac.in\/jee-main-syllabus\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">JEE Main 2022 Syllabus<\/span><\/a><span style=\"font-weight: 400;\"> and <\/span><a href=\"https:\/\/www.aakash.ac.in\/jee-advanced-syllabus\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">JEE Advanced 2022 Syllabus.<\/span><\/a><\/p>\n<h3>Key Terminologies in Coordinate Geometry<\/h3>\n<p><span style=\"font-weight: 400;\">Some of the crucial terms interconnected with coordinate geometry that the students must clearly understand. Those crucial terminologies are:<\/span><\/p>\n<ul>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Coordinates: A set of values that helps to ascertain and represent the exact position of a point in the coordinate plane are known as Coordinates.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Coordinate Plane: A Coordinate Plane is a 2D plane. It is formed by the intersection of two perpendicular lines which are called the x-axis and y-axis.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Abscissa: Abscissa or x-coordinate is the distance of any point from the y-axis.<\/span><\/li>\n<li style=\"font-weight: 400;\" aria-level=\"1\"><span style=\"font-weight: 400;\">Ordinate: Ordinate or y-coordinate is the distance of a point from the x-axis.<\/span><\/li>\n<\/ul>\n<p><span style=\"font-weight: 400;\">Do not miss to grab detailed explanations on various topics and master your subjects without any difficulty. Study <\/span><a href=\"https:\/\/www.aakash.ac.in\/important-concepts\/maths\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">Maths Concepts<\/span><\/a><span style=\"font-weight: 400;\"> explained by professionals and explore them now!\u00a0<\/span><\/p>\n<p><b>Coordinate Geometry: Explanatory Notes<\/b><\/p>\n<p><b>Understanding Coordinate and A Coordinate Plane<\/b><\/p>\n<p><span style=\"font-weight: 400;\">The students must be aware of plotting graphs on a plane. We are given values or numbers in a table form for both linear and nonlinear equations. Here, the number line which is also called a Cartesian plane features four quadrants formed by two intersecting axes which are perpendicular to each other. The horizontal line is represented as the x-axis and the vertical line is represented as the y-axis.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">The four quadrants created via the division of the Cartesian plane corresponding to their respective values are:<\/span><\/p>\n<table>\n<tbody>\n<tr>\n<td><span style=\"font-weight: 400;\">Quadrant I<\/span><\/td>\n<td><span style=\"font-weight: 400;\">Positive x and Positive y<\/span><\/td>\n<\/tr>\n<tr>\n<td><span style=\"font-weight: 400;\">Quadrant II<\/span><\/td>\n<td><span style=\"font-weight: 400;\">Negative x and Positive y<\/span><\/td>\n<\/tr>\n<tr>\n<td><span style=\"font-weight: 400;\">Quadrant III<\/span><\/td>\n<td><span style=\"font-weight: 400;\">Negative x and Negative y<\/span><\/td>\n<\/tr>\n<tr>\n<td><span style=\"font-weight: 400;\">Quadrant IV<\/span><\/td>\n<td><span style=\"font-weight: 400;\">Positive x and Negative y<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><span style=\"font-weight: 400;\">Both the axes intersect at a point, which is known as the Origin (o). The location of any point on the Cartesian plane is indicated by a pair of values, called Coordinates. They are represented as x and y.<\/span><\/p>\n<table>\n<tbody>\n<tr>\n<td><b>Key Takeaway<\/b><\/p>\n<p><span style=\"font-weight: 400;\">The four quadrants are labelled counterclockwise. The first quadrant occupies the upper right-hand portion of the Cartesian plane.<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><span style=\"font-weight: 400;\">Once we identify the coordinates, then we can easily calculate the distance between the two points. Also, we can find the midpoint of that interval, which connects the points.<\/span><\/p>\n<p><b>Representation of Equation of a Line on a Cartesian Plane<\/b><\/p>\n<p><span style=\"font-weight: 400;\">The equation of a line can be represented in several ways.\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">We represent a general form of line as:\u00a0<\/span><\/p>\n<p><b>Ax + By + C = 0.<\/b><\/p>\n<p><b>Slope intercept Form<\/b><span style=\"font-weight: 400;\">\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">To represent this way, let x &amp; y be the coordinate of a point, through which a line passes. Now, m be the slope of a line, and c be the y-intercept. Then the equation of a line can be represented as :<\/span><\/p>\n<p><span style=\"font-weight: 400;\">y = mx + c<\/span><\/p>\n<p><b>Slope of a Line<\/b><\/p>\n<p><span style=\"font-weight: 400;\">The general form of a line is Ax + By + C = 0. Now, the slope of this line can be found by converting this general form into the slope-intercept form.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Ax + By + C = 0\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u21d2 By = \u2212 Ax \u2013 C<\/span><\/p>\n<p><span style=\"font-weight: 400;\">or,<\/span><\/p>\n<p><span style=\"font-weight: 400;\">y = -(Ax\/B) &#8211; (C\/B)<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Here, on comparing the above-formed equation with the slope-intercept form, i.e., y = mx + c, we get,<\/span><\/p>\n<p><b>m = -A\/B<\/b><\/p>\n<p><span style=\"font-weight: 400;\">Hence, we can find the slope of a line directly from the general equation of a line.<\/span><\/p>\n<p><b>Formulas and Theorems in Coordinate Geometry<\/b><\/p>\n<p><span style=\"font-weight: 400;\">Here are some essential formulas covered in class 10 maths Coordinate Geometry &#8211;<\/span><\/p>\n<p><b>Distance Formula :<\/b><span style=\"font-weight: 400;\">\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Let the two points be A and B. These have coordinates (x<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">,y<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">) and (x<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">,y<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">), respectively.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">The distance formula is:\u00a0\u00a0<\/span><\/p>\n<p><b>d = \u221a((x<\/b><b>2<\/b><b> &#8211; x<\/b><b>1<\/b><b>)2+(y<\/b><b>2<\/b><b>-y<\/b><b>1<\/b><b>)2)<\/b><\/p>\n<p><span style=\"font-weight: 400;\">Students can use this to find the distance between two points.<\/span><\/p>\n<p><b>Mid-point Theorem<\/b><\/p>\n<p><span style=\"font-weight: 400;\">It is useful to find the midpoint of a line connecting two points.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Again assume two points A and B, having coordinates (x<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">,y<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">) and (x<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">,y<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">) respectively. Now, let M(x,y) be the midpoint lying on the line connecting points A and B. The pair of coordinates of the point M will be:<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u00a0<\/span><b>M(x,y) = [(x<\/b><b>1<\/b><b>+x<\/b><b>2<\/b><b>)\/2, (y<\/b><b>1<\/b><b>+y<\/b><b>2<\/b><b>)\/2]<\/b><\/p>\n<p><b>Section Formula<\/b><\/p>\n<p><span style=\"font-weight: 400;\">Students can use this to find a point that divides a line into an l:m ratio.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Let the line AB have coordinates (x<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">,y<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">) and (x<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">,y<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">). Now, assume D as a point that divides the line in the ratio l:m. Then the coordinates of the point P are given as-<\/span><\/p>\n<p><span style=\"font-weight: 400;\">If the ratio l:m is internal:<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u00a0<\/span><b>[(lx<\/b><b>2<\/b><b>+mx<\/b><b>1<\/b><b>)\/l+m, (ly<\/b><b>2<\/b><b>+my<\/b><b>1<\/b><b>)\/l+m]<\/b><\/p>\n<p><span style=\"font-weight: 400;\">If the ratio l:m is external:<\/span><\/p>\n<p><b>[(lx<\/b><b>2<\/b><b>-mx<\/b><b>1<\/b><b>)\/l-m, (ly<\/b><b>2<\/b><b>-my<\/b><b>1<\/b><b>)\/l-m]<\/b><\/p>\n<p><b>Area of a Triangle in Cartesian Plane<\/b><\/p>\n<p><span style=\"font-weight: 400;\">In coordinate geometry, students can find the area of a triangle by this formula.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Let (x<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">,y<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">), (x<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">,y<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">), and (x<\/span><span style=\"font-weight: 400;\">3<\/span><span style=\"font-weight: 400;\">,y<\/span><span style=\"font-weight: 400;\">3<\/span><span style=\"font-weight: 400;\">) be the three vertices. Now, area of triangle =<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u00a0<\/span><b>1\/2 * [ x<\/b><b>1<\/b><b>(y<\/b><b>2<\/b><b> &#8211; y<\/b><b>3<\/b><b>) + x<\/b><b>2<\/b><b>(y<\/b><b>3<\/b><b> &#8211; y<\/b><b>1<\/b><b>) + x<\/b><b>3<\/b><b>(y<\/b><b>1<\/b><b> &#8211; y<\/b><b>2<\/b><b>) ]<\/b><\/p>\n<p><span style=\"font-weight: 400;\">The three points are collinear if the area of a triangle with vertices are (x1,y1), (x2,y2), and (x3,y3) is zero.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Now, it is time for some examples. Consider a small exercise below.<\/span><\/p>\n<p><strong>1. Find the distance between the two points A(1, 2) and B(-2, 3).<\/strong><\/p>\n<p><span style=\"font-weight: 400;\">Sol: Given two points, J (1, 2) and K (-2, 3).<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Let J(1, 2) = (x<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">, y<\/span><span style=\"font-weight: 400;\">1<\/span><span style=\"font-weight: 400;\">) and K(-2, 3) = (x<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">, y<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">)<\/span><\/p>\n<p><span style=\"font-weight: 400;\">To calculate the distance between J and K using the distance formula,<\/span><\/p>\n<p><span style=\"font-weight: 400;\">D = \u221a[(x2 &#8211; x1)<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\"> + (y2 &#8211; y1)<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">]<\/span><\/p>\n<p><span style=\"font-weight: 400;\">JK = \u221a[(-2 &#8211; 1)<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\"> + (3 &#8211; 2)<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\">]<\/span><\/p>\n<p><span style=\"font-weight: 400;\">JK = \u221a(-3)<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\"> + (1)<\/span><span style=\"font-weight: 400;\">2<\/span><\/p>\n<p><span style=\"font-weight: 400;\">JK = \u221a9+1 = 3\u221a1 unit.<\/span><\/p>\n<p><strong>2. Find the equation of the straight line passing through (2, 3) and is perpendicular to the line 3x + 2y + 4 = 0.<\/strong><\/p>\n<p><span style=\"font-weight: 400;\">Sol: The given line is 3x + 2y + 4 = 0\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">=&gt; y = -3x \/ 2 &#8211; 2<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Any line perpendicular to it will have slope = 2\/3<\/span><\/p>\n<p><span style=\"font-weight: 400;\">So, equation of line through (2, 3) and slope 2 \/ 3 will be &#8211;\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">(y \u2013 3) = 2 \/ 3 (x \u2013 2)<\/span><\/p>\n<p><span style=\"font-weight: 400;\">=&gt; 3y \u2013 9 = 2x \u2013 4<\/span><\/p>\n<p><span style=\"font-weight: 400;\">=&gt; 3y \u2013 2x \u2013 5 = 0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Hence, the equation is 3y \u2013 2x \u2013 5 = 0.<\/span><\/p>\n<p><strong>3. Find the area of the triangle formed by the vertices (5,6), (2,4), and (1,-3).<\/strong><\/p>\n<p><span style=\"font-weight: 400;\">Sol: 1\/2* [5(4 &#8211; (-3)) + 2(-3 &#8211; 6) + 1(6 &#8211; 4)]<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01\/2* [5(7) +2(-9) + 1(2)]<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01\/2* [35 &#8211; 18 + 2]<\/span><\/p>\n<p><span style=\"font-weight: 400;\">\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01\/2* [19] = 9.5<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Hence, the answer is 9.5 square units.<\/span><\/p>\n<p><strong>4. At what point does the line 3x + y = -6 intercept the x-axis?<\/strong><\/p>\n<p><span style=\"font-weight: 400;\">Sol: The line 3x + y = -6 will intercept x-axis at y = 0.\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Substituting this value of y in the above equation.\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">3x + 0 = -6<\/span><\/p>\n<p><span style=\"font-weight: 400;\">x = -6\/3 = -2<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Hence, the line will intercept the x-axis at (-2,0).<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Get Expert Guidance for <\/span><a href=\"https:\/\/www.aakash.ac.in\/jee-main-exam\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">JEE Main 2022 Exam<\/span><\/a><span style=\"font-weight: 400;\"> Pattern and <\/span><a href=\"https:\/\/www.aakash.ac.in\/jee-main-counselling\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">JEE Main Counselling<\/span><\/a><span style=\"font-weight: 400;\">.<\/span><\/p>\n<h3>Conclusion<\/h3>\n<p><span style=\"font-weight: 400;\">Coordinate Geometry is that branch of geometry where we determine the position of a point using coordinates. Coordinate Geometry can be a slightly confusing topic for some students because it involves formulas and algebraic calculations in the cartesian plane. However, we have explained the vital concepts of this topic here. Students can get full support, including study notes and doubt clearing sessions at our platform. We suggest that students should do regular practice applying these formulas to sums to gain a good command of the topic. Our notes can be of great help to recall the concepts whenever required.\u00a0<\/span><\/p>\n<h2>FAQs<\/h2>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Class 10 Mathematics Exercise Notes for Chapter 7 &#8211; Coordinate Geometry describes the various core concepts of coordinate geometry. Coordinate Geometry helps the students to interpret and quantitatively present geometrical shapes. It helps to extract numerical information by taking out logical conclusions. We have Class 10 Mathematics Exercise Notes curated by Expert Teachers. 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