source: ncert
topic: geometry
class_level: class_11
chapter: ch10
difficulty: intermediate
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.