Question

What is the equation of a line that passes through the point (8, 1) and is perpendicular to the line whose equation is y=−2/3x+5y=−2/3x+5 ?

Answer 1

y = -2/3x + 5....the slope here is -2/3. A perpendicular line will have a negative reciprocal slope. To find the negative reciprocal, we flip the slope and change the sign. So our perpendicular line will have a slope of 3/2 (see how I flipped -2/3 making it -3/2....and then changed the sign, making it 3/2)

y = mx + b
slope(m) = 3/2
(8,1)....x = 8 and y = 1
now we sub and find b, the y int
1 = 3/2(8) + b
1 = 12 + b
1 - 12 = b
-11 = b

so ur perpendicular equation is : y = 3/2x - 11 <==

Answer 2

your equation oinitial is y=-2/3x+5

perpendicular lines have slopes that mulitply to get -1

y=mx+b
m=slope

given
y=-2/3x+5
slope=-2/3

perpendicular so
-2/3 times what=-1
times -3/2 both sides
what=3/2
the slope is 3/2

let's use point slope form

equation of a line that passes through (x1,y1) and has a slope of m is
y-y1=m(x-x1)
so given point (8,1) and slope 3/2

y-1=3/2(x-8)
is yo equation
if yo wanted slope intercept
y-1=3/2x-12
y=3/2x-11 is slope intercept
if wanted standard form

11=3/2x-y
3x-2y=22


equaiton is
$y-1=\frac{3}{2}(x-8)$ for point slope form
$y=\frac{3}{2}x-11$ in slope intercept form
3x-2y=22 in standard form