Question

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

Answer 1

y-7=-(4/3)(x-5) (point slope form)

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

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

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