Find the different forms of the equation of the straight line passing through ( 5 , 7 ) and perpendicular to the straight line whose equation is 4y-3x+9=0

y-7=-(4/3)(x-5)  (point slope form)

y=-(4/3)x+(41/3) (slope intercept form)

4x+3y-41=0 (general form)

Step-by-step explanation:

step 1

Find the slope of the perpendicular line to 4y-3x+9=0

If two lines are perpendicular, then the product of their slopes is equal to -1

4y-3x+9=0

4y=3x-9

y=(3/4)x-9/4 > the slope is m=3/4

The perpendicular slope is

m*(3/4)=-1

m=-4/3

step 2

Find the equation of the line into slope point form

y-y1=m(x-x1)

we have

(x1,y1)=(5,7)

m=-4/3

substitute

y-7=-(4/3)(x-5) > equation of the line into point slope form

Convert to slope intercept form > y=mx+b

y-7=-(4/3)(x-5)

y-7=-(4/3)x+(20/3)

y=-(4/3)x+(20/3)+7

y=-(4/3)x+(41/3) > equation of the line into slope intercept form

Convert to general form > Ax+By+C=0

y=-(4/3)x+(41/3)

Multiply by 3 both sides

3y=-4x+41

4x+3y-41=0 > equation of the line in general form

center (1,3) with radius 6

step-by-step explanation:

we must complete the square to find the center and radius of the circle.

first make sure the x and y squared terms have 1 as their coefficients. we also make sure x and y terms together.

we now create space between the x and y terms with parenthesis.

we complete the square by taking the middle terms -2x and the -6y - divide each and square them.

we add the squares to both sides.

simplify.

and write the quadratics in factored form.

the center is (h,k) or (1,3). the radius is the square root of 36 which is 6.

sorry

step-by-step explanation:

sorry, but im pretty sure its a now that i look at it