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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:

Thus, the point-slope form of line

â‡’

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

Consider a line L with slope

Thus, the point-slope form of line

â‡’

â‡’

This equation is the Slope-Intercept Form, where:

Consider a line

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 (

Two-Point Form:

Thus, the two-point form of line

â‡’

â‡’

Dividing both sides by

(

â‡’

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

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‰ = $\frac{\text{OM}}{\text{OA}}$

â‡’ Cos Ï‰ = $\frac{\text{OM}}{\text{P}}$

â‡’ OM =

Sin Ï‰ = $\frac{\text{AM}}{\text{OA}}$

â‡’ Sin Ï‰ = $\frac{\text{AM}}{\text{P}}$

â‡’ AM =

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¥

Slope of

Slope of OA = tan Ï‰

âˆ´

â‡’

â‡’

â‡’

Point-Slope Form:

Thus, the point-slope form of line

â‡’

â‡’

Normal Form:

where:

Ï‰ = Inclination of the normal

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:

Thus, the point-slope form of line

â‡’

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

Consider a line L with slope

Thus, the point-slope form of line

â‡’

â‡’

This equation is the Slope-Intercept Form, where:

Consider a line

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 (

Two-Point Form:

Thus, the two-point form of line

â‡’

â‡’

Dividing both sides by

(

â‡’

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

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‰ = $\frac{\text{OM}}{\text{OA}}$

â‡’ Cos Ï‰ = $\frac{\text{OM}}{\text{P}}$

â‡’ OM =

Sin Ï‰ = $\frac{\text{AM}}{\text{OA}}$

â‡’ Sin Ï‰ = $\frac{\text{AM}}{\text{P}}$

â‡’ AM =

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¥

Slope of

Slope of OA = tan Ï‰

âˆ´

â‡’

â‡’

â‡’

Point-Slope Form:

Thus, the point-slope form of line

â‡’

â‡’

Normal Form:

where:

Ï‰ = Inclination of the normal