Perpendicular distance from a point to a line — A/L Combined Mathematics

What must I remember about Perpendicular distance from a point to a line?

• Perpendicular distance d from point P(x1, y1) to straight line L: ax + by + c = 0 is d = |ax1 + by1 + c| / √(a² + b²).
• Derivation via area of triangle: Form right-angled triangle with intercepts or use parametric normal projection: d = |ax1 + by1 + c| / √(a² + b²).
• Coordinates of the Foot of the Perpendicular (x0, y0) from P(x1, y1) to ax + by + c = 0: (x0 - x1) / a = (y0 - y1) / b = - (ax1 + by1 + c) / (a² + b²).
• Coordinates of the Image (Reflection) P'(x', y') of point P(x1, y1) across line ax + by + c = 0: (x' - x1) / a = (y' - y1) / b = - 2(ax1 + by1 + c) / (a² + b²).
• Equations of Angle Bisectors between lines a1*x + b1*y + c1 = 0 and a2*x + b2*y + c2 = 0: (a1*x + b1*y + c1) / √(a1² + b1²) = ± (a2*x + b2*y + c2) / √(a2² + b2²).

Test yourself (1 minute)

1. Distinguishing Acute from Obtuse angle bisector

   Make constant terms c1, c2 positive; compute (a1*a2 + b1*b2): if a1*a2 + b1*b2 > 0, the '+' sign yields obtuse angle bisector and '-' yields acute angle bisector; if a1*a2 + b1*b2 < 0, '+' yields acute and '-' yields obtuse.

2. Angle bisector containing origin

   When c1, c2 > 0, the bisector corresponding to the '+' sign contains the origin.

3. Perpendicular distance from Origin

   Distance from O(0, 0) to ax + by + c = 0 simplifies directly to d = |c| / √(a² + b²).

Practise Perpendicular distance from a point to a line questions today

Source: NIE G.C.E. (A/L) Combined Mathematics Resource Book - Pure Mathematics, Unit 18

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Year 1 · Term 3