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Straight Line — Class 11 Maharashtra HSC Mathematics revision notes

A 4-page visual revision sheet for Straight Line, written for Class 11 Maharashtra HSC Mathematics. Formulas, diagrams and the traps that cost marks, laid out to be read the night before rather than the term before. Free, and it opens inside CapCompass.

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Straight Line in one look

a locus is a set of points obeying a rule, and a line is the simplest set whose slope never changes

Locus and the Equation of a Locus

  • a locus is the set of all points satisfying one geometric condition, plural loci
  • an equation OF a locus holds on every point of it, none off it
  • method: let P(x,y) be general, translate the condition into an x,y equation, simplify
  • e.g. perpendicular bisector of AB: locus of P with PA=PB, a line
  • e.g. bisector of an angle: locus of points equidistant from its arms, a ray

Q. Find the equation of the locus of points equidistant from A(-3,0) and B(3,0). Identify the locus.

  1. let P(x,y) satisfy PA=PB, so PA^2=PB^2
  2. (x+3)^2+y^2=(x-3)^2+y^2
  3. x^2+6x+9=x^2-6x+9, so 12x=0

Answer: x=0: the locus is the Y-axis, the perpendicular bisector of AB

Shift of Origin

  • shift the origin to O'(h,k), axes kept parallel: x=x'+h, y=y'+k
  • to move a locus into the new axes, substitute x'+h for x, y'+k for y, simplify

Q. The origin is shifted to (2,3), axes parallel. Find the new equation of x^2-xy-2y^2-x+4y+2=0

  1. substitute x=X+2, y=Y+3
  2. (X+2)^2-(X+2)(Y+3)-2(Y+3)^2-(X+2)+4(Y+3)+2=0
  3. expand and collect every constant and cross term

Answer: X^2-XY-2Y^2-10Y-8=0

Inclination and Slope

  • inclination θ: smallest angle a line makes with the positive X-axis, anticlockwise; 0°≤θ<180°
  • slope m=tanθ when it exists; undefined for the Y-axis and lines parallel to it
  • slope of the X-axis, and of lines parallel to it, is 0
  • parallel lines share the same inclination, equivalently the same slope
  • two points A(x_1,y_1), B(x_2,y_2) on a non-vertical line: m=y_2-y_1/x_2-x_1

Perpendicular and Angle Between Lines

  • perpendicular m_1m_2=-1
  • acute angle between two lines tanθ=left|m_1-m_2/1+m_1m_2right|

Q. Find the acute angle between lines of slope 3 and -2

  1. tanθ=left|3-(-2)/1+(3)(-2)right|=left|5/-5right|=1

Answer: tanθ=1, so θ=45°

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