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# Equations of Lines (Part II)

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#### Equations of Lines (Part II) - Lesson Summary

Slope-Intercept Form

Consider a line L with slope m that intersects the Y-axis at a point A, which is at distance C from the origin. That is the given line passes through the point zero, C.

Point-Slope Form: yy 1 = m( xx 1)

Thus, the point-slope form of line l  is: yc = m( x – 0)

y = mx + c

This equation is the Slope-Intercept Form of the line, where:

m = Slope of the line

c = y-intercept of the line

Consider a line L with slope m, whose x-intercept is given as D. The given line passes through the point B ( d, 0).

Thus, the point-slope form of line l:

yy 1 = m( xx 1)

y – 0 = m( xd)

y = m( xd)

This equation is the Slope-Intercept Form, where:

m = Slope of the line

d = x-intercept of the line

Intercept  Form

Consider a line l with x-intercept A and y-intercept B.

Let’s denote the points where the line intersects the X-axis and the Y-axis as P and Q, respectively.

Therefore, the coordinates of point ‘P’ are ( a, 0) and that of point ‘Q’ are (0, b).

Two-Point Form: yy 1 = {( y 2y 1)/ ( x 2x 1)} x ( xx 1)

Thus, the two-point form of line l is: y -  0 = {( b – 0)/ (0 – a)} x ( xa)

ay = - bx + ab

bx + ay = ab

Dividing both sides by ab:

( bx + ay)/ ab = ab/ ab

x/ a + y/ b = 1

The above equation is the Intercept Form of equation of the straight line.

Normal Form

Consider a non-vertical line L. Let OA be the normal or perpendicular drawn to line L from the origin.

Let the perpendicular distance of line L from the origin, or the length of perpendicular OA, be P.

∠AOX = ω

AM ⊥ OX

Distance OM represents the x-coordinate, and distance AM represents the y-coordinate of point A.

From right angled triangle OMA

Cos ω = OM OA

⇒ Cos ω = OM P

⇒ OM = P Cos ω

Sin ω = AM OA

⇒ Sin ω = AM P

⇒ AM = P Sin ω

For the given values of P and omega, the coordinates of point A will remain the same no matter the quadrant in which line L lies.

The coordinates of a point on line L, we can find its equation if we know its slope, using the point-slope form. Let M be the slope of the given line L.

Given OA ⊥ l:

Slope of l x slope of OA = -1

Slope of OA = tan ω

m tan ω = -1

m = -1/ tan ω

m = -cot ω

m = -cos ω/ sin ω

Point-Slope Form: yy 1 = m( xx 1)

Thus, the point-slope form of line l is: yp sin ω = (-cos ω/ sin ω)( xp cos ω)

y sin ω – p sin 2 ω = - x cos ω + p cos 2 ω)

x cos ω + y sin ω = p(sin 2 ω + cos 2 ω)⇒ x cos ω + y sin ω = p     [ ∵ sin 2 ω + cos 2 ω = 1]

Normal Form: x cos ω + y sin ω = p

where:

p = normal distance of the line from origin

ω = Inclination of the normal