Reduce the line `2x-3y+5=0` in slope-intercept, intercept, and normal forms.

Equation of given line is
2x-3y+5=0
` or y = (2x)/(3) + (5)/(3)`
This is of the form y=mx +c.
Slope of the line is m = tan` theta = (2)/(3) " and y-intercept is c" = (5)/(3)`.
Here, `theta` is the angle of the line with positive x-axis.
Intercept form of the line is `(x)/((-(5)/(2))) + (y)/(((5)/(3))) = 1`, where x - intercept
`"is a "= -(5)/(2) "and y-intercept is b"= (5)/(3).`
Furthe to get the equation in normal form, divide both sides by
`sqrt(2^(2) + 3^(2)) = sqrt(13).`
`"So, equation of line becomes" - (2x)/(sqrt(13)) + (3y)/(sqrt(13)) = (5)/(sqrt(13)).`
Comparing with standard normal form x cos `alpha` + y sin `alpha` = p,
we get
` "cos" alpha = -(2)/(sqrt(13)), "sin" alpha = (3)/(sqrt(13)) and`
`p = (5)/(sqrt(13)) = "Distance of line fromo origin"`
Also, inclination of normal to the line with x-axis is
`alpha = 180^(@) - "sin"^(-1) (3)/(sqrt(13)).`