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

not enough information

step-by-step explanation:

we cannot know this, because the other students could have handed out 10 flyers, or they could have handed out 100. one way, the mean would go up, and another way, the mean would go down.

15 units use socratic

step-by-step explanation: 15 units