| --- |
| source: ncert |
| topic: geometry |
| class_level: class_10 |
| chapter: ch7 |
| difficulty: intermediate |
| --- |
| |
| Class 10 | Ch7: Coordinate Geometry |
| DISTANCE FORMULA: |
| Distance between P(x1,y1) and Q(x2,y2) = sqrt((x2-x1)^2+(y2-y1)^2). |
| Distance from origin: OP=sqrt(x^2+y^2). |
| SECTION FORMULA: |
| Point P dividing AB (A(x1,y1), B(x2,y2)) in ratio m:n INTERNALLY: |
| P = ((mx2+nx1)/(m+n), (my2+ny1)/(m+n)). |
| MIDPOINT FORMULA (m=n=1): |
| M = ((x1+x2)/2, (y1+y2)/2). |
| EXTERNAL DIVISION: |
| P = ((mx2-nx1)/(m-n), (my2-ny1)/(m-n)). |
| AREA OF TRIANGLE: |
| Given A(x1,y1), B(x2,y2), C(x3,y3): |
| Area = (1/2)|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|. |
| If area=0, points are COLLINEAR. |
| CENTROID OF TRIANGLE: |
| G = ((x1+x2+x3)/3, (y1+y2+y3)/3). Divides each median in 2:1. |
| SOLVED EXAMPLES: |
| Example 1: Distance between A(2,3) and B(-1,0). |
| AB=sqrt((-1-2)^2+(0-3)^2)=sqrt(9+9)=sqrt(18)=3sqrt(2). |
| Example 2: Find point dividing (1,3) and (4,6) in ratio 2:1 internally. |
| P=((2×4+1×1)/(2+1),(2×6+1×3)/(2+1))=(9/3,15/3)=(3,5). |
| Example 3: Area of triangle A(2,3), B(-1,0), C(2,-4). |
| Area=(1/2)|2(0-(-4))+(-1)((-4)-3)+2(3-0)|=(1/2)|8+7+6|=(1/2)(21)=10.5 sq units. |
| Example 4: Show A(1,2), B(3,4), C(5,6) are collinear. |
| Area=(1/2)|1(4-6)+3(6-2)+5(2-4)|=(1/2)|(-2)+12+(-10)|=(1/2)(0)=0. Collinear. |
| COMMON MISTAKES: |
| Section formula: m corresponds to the second point B, n to first point A. |
| Area formula: use ABSOLUTE VALUE (modulus bars) — area cannot be negative. |
| Midpoint: average of BOTH x-coordinates and BOTH y-coordinates separately. |
