| --- |
| source: ncert |
| topic: geometry |
| class_level: class_9 |
| chapter: ch3 |
| difficulty: beginner |
| --- |
| |
| Class 9 | Ch3: Coordinate Geometry |
| CARTESIAN PLANE: |
| Two perpendicular lines: x-axis (horizontal) and y-axis (vertical). |
| Origin O = (0,0). Point P = (x,y) where x=abscissa, y=ordinate. |
| QUADRANTS: |
| I: (+,+) — right of y-axis, above x-axis. |
| II: (-,+) — left of y-axis, above x-axis. |
| III: (-,-) — left of y-axis, below x-axis. |
| IV: (+,-) — right of y-axis, below x-axis. |
| Points ON axes: x-axis has y=0. y-axis has x=0. |
| PLOTTING POINTS: |
| Start at origin. Move right/left for x (positive=right, negative=left). |
| Then move up/down for y (positive=up, negative=down). Mark the point. |
| DISTANCE FORMULA (introduced here, developed in Class 10): |
| Distance between A(x1,y1) and B(x2,y2) = sqrt((x2-x1)^2+(y2-y1)^2). |
| COLLINEAR POINTS: three points lie on same straight line. |
| SECTION FORMULA (for later): divides line segment in given ratio. |
| GRAPHS OF LINEAR EQUATIONS: |
| y=x: line through origin at 45°. y=-x: line through origin at 135°. |
| x=a: vertical line. y=b: horizontal line. |
| SOLVED EXAMPLES: |
| Example 1: Plot A(3,4), B(-2,3), C(-4,-1), D(2,-3). |
| A: Quadrant I (both positive). B: Quadrant II. C: Quadrant III. D: Quadrant IV. |
| Example 2: Find distance between A(3,4) and B(-2,3). |
| AB = sqrt((-2-3)^2+(3-4)^2) = sqrt(25+1) = sqrt(26) units. |
| Example 3: In which quadrant is (-3,-5)? |
| Both negative → Quadrant III. |
| COMMON MISTAKES: |
| Point (3,4) is NOT same as (4,3). First coordinate is always x (horizontal). |
| Origin (0,0) is not in any quadrant — it's at the intersection of axes. |
| Points on x-axis have y-coordinate=0, not x-coordinate=0. |
