Use the given conditions to write an equation for the line in point-slope form and general form. passing through (-9,9) and parallel to the line whose equation is 2x-5y-3=0. the equation of the line in point slope form is the equation of the line in general form is = 0. asap.

The equation of a parallel line through point (h, k) can be found by altering the given line's equation this way:

To change this equation to point-slope form, solve for y (without expanding the x-term). If there is any constant added to the x-term, add its opposite (so it appears with y).

Here, that looks like ...

... 2(x+9) -5(y-9) = 0

... -5(y-9) = -2(x+9) . . . . add the opposite of the x-term

... y - 9 = 2/5(x +9) . . . . this is the point-slope form you want

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To put the equation in general form, multiply by the common denominator of any fractions, and subtract the stuff on the right side of the equal sign. Factor out any common factors of the integer coefficients and constant, and multiply by -1 if the leading coefficient is negative.*

Here, we multiply by 5 (and eliminate parentheses) to get

... 5y -45 = 2x +18

... -2x +5y -63 = 0 . . . . right side subtracted

... 2x -5y +63 = 0 . . . . . leading coefficient made positive. This is your general form.

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* Actually, I can find no requirement that General Form have a positive leading coefficient. Nor is there a requirement that the coefficients be mutually prime. However, these are requirements for the very similar Standard Form, which has the constant on the right side of the equal sign, instead of the left. Since leading minus signs tend to get lost, IMO, it is a good idea to make the leading coefficient positive, even if it is not a requirement.

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)