| --- |
| source: ncert |
| topic: geometry |
| class_level: class_11 |
| chapter: ch10 |
| difficulty: intermediate |
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| |
| Class 11 | Ch10: Straight Lines |
| SLOPE: m=tan(theta)=(y2-y1)/(x2-x1). Horizontal line: m=0. Vertical line: undefined. |
| Parallel lines: m1=m2. Perpendicular lines: m1×m2=-1. |
| EQUATIONS OF LINES: |
| Slope-intercept: y=mx+c (m=slope, c=y-intercept). |
| Point-slope: y-y1=m(x-x1). |
| Two-point form: (y-y1)/(y2-y1)=(x-x1)/(x2-x1). |
| Intercept form: x/a+y/b=1 (a=x-intercept, b=y-intercept). |
| Normal form: x cos(alpha)+y sin(alpha)=p (p=perpendicular distance from origin, alpha=angle). |
| General form: ax+by+c=0. |
| DISTANCE FORMULAS: |
| Distance from point (x1,y1) to line ax+by+c=0: d=|ax1+by1+c|/sqrt(a^2+b^2). |
| Distance between parallel lines ax+by+c1=0 and ax+by+c2=0: |c1-c2|/sqrt(a^2+b^2). |
| ANGLE BETWEEN LINES: |
| tan(theta)=|(m1-m2)/(1+m1m2)|. |
| SOLVED EXAMPLES: |
| Example 1: Line through (2,-3) perpendicular to 3x-4y+5=0. |
| Slope of given line=3/4. Perpendicular slope=-4/3. |
| Line: y-(-3)=(-4/3)(x-2) → 3y+9=-4x+8 → 4x+3y+1=0. |
| Example 2: Distance between 3x+4y-5=0 and 3x+4y+15=0. |
| d=|(-5)-(15)|/sqrt(9+16)=20/5=4 units. |
| Example 3: Find foot of perpendicular from (1,2) to line 2x-3y+4=0. |
| Use formula: (x-x1)/a=(y-y1)/b=-(ax1+by1+c)/(a^2+b^2). |
| =(x-1)/2=(y-2)/(-3)=-(2-6+4)/(4+9)=0/13=0. |
| So x=1, y=2. Point itself lies on line. (Check: 2-6+4=0. Yes.) |
| COMMON MISTAKES: |
| Perpendicular lines: m1×m2=-1 (product is -1, not sum). |
| Distance formula: divide by sqrt(a^2+b^2) NOT (a^2+b^2). |
| Parallel lines have SAME slope. Perpendicular lines: slopes are NEGATIVE RECIPROCALS. |
