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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: y â€“ y 1 = m( x â€“ x 1)

Thus, the point-slope form of line l  is: y â€“ c = m( x â€“ 0)

â‡’ 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:

y â€“ y 1 = m( x â€“ x 1)

â‡’ y â€“ 0 = m( x â€“ d)

â‡’ y = m( x â€“ d)

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.

Letâ€™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 â€˜Pâ€™ are ( a, 0) and that of point â€˜Qâ€™ are (0, b).

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

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

â‡’ ay = - bx + ab

â‡’ bx + ay = ab

Dividing both sides by ab:

( bx + ay)/ ab = ab/ ab

â‡’ 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.

âˆ AOX = Ï‰

AM âŠ¥ OX

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

From right angled triangle OMA

Cos Ï‰ = OM OA

â‡’ Cos Ï‰ = OM P

â‡’ OM = P Cos Ï‰

Sin Ï‰ = AM OA

â‡’ Sin Ï‰ = AM P

â‡’ AM = P Sin Ï‰

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 âŠ¥ l:

Slope of l x slope of OA = -1

Slope of OA = tan Ï‰

âˆ´ m tan Ï‰ = -1

â‡’ m = -1/ tan Ï‰

â‡’ m = -cot Ï‰

â‡’ m = -cos Ï‰/ sin Ï‰

Point-Slope Form: y â€“ y 1 = m( x â€“ x 1)

Thus, the point-slope form of line l is: y â€“ p sin Ï‰ = (-cos Ï‰/ sin Ï‰)( x â€“ p cos Ï‰)

â‡’ y sin Ï‰ â€“ p sin 2 Ï‰ = - x cos Ï‰ + p cos 2 Ï‰)

â‡’ x cos Ï‰ + y sin Ï‰ = p(sin 2 Ï‰ + cos 2 Ï‰)â‡’ x cos Ï‰ + y sin Ï‰ = p     [ âˆµ sin 2 Ï‰ + cos 2 Ï‰ = 1]

Normal Form: x cos Ï‰ + y sin Ï‰ = p

where:

p = normal distance of the line from origin

Ï‰ = Inclination of the normal