Cordinate Plane Activity?

Question

Given the line 2x – 3y = 9 and the point (4, –1), find lines through the point that are

(a) parallel to the given line and
(b) perpendicular to it.

How do I do this? I need details

Answer 1

2x – 3y = 9; solve for y
-3y = -2x + 9
y = 2/3 x - 3

the slope is 2/3

a parallel line will have the same slope, so plug (4, -1) into
y = 2/3x + b, and solve for b

-1 = 2/3 (4) + b
-1 = 2 2/3 + b
-3 2/3 = b

y = 2/3 x - 3 2/3

a perpendicular line has a slope that is the negative reciprocal of 2/3 which is -3/2
y = -3/2 x + b
-1 = -3/2 (4) + b
-1 = -6 + b
5 = b

y = -3/2 x + 5

Answer 2

Put it in y=mx+b form. 2x-9=3y, y=(2/3)x-3. So the slope is 2/3.

A line parallel to that line also has slope 2/3. So, to find the line of slope 2/3 through (4,-1), we use y=mx+b again. -1=(2/3)(4)+b => -1=11/3 + b, => b=-14/3. So y=(2/3)x-14/3.

The perpendicular line has the negative reciprocal slope, so -3/2. We do the same thing again: -1=(-3/2)(4)+b, => -1=-6+b, b=-7.

Answer 3

To write equation for the line you need:1) slope, 2) point
you have the point (4,-1), but need slope
so solve for y in the given equation, and identify slope
2x - 3y = 9
-3y = 9 - 2x
y = -3 + (2/3) x, so m (slope) = 2/3

Now for parallel line you keep the same slope: 2/3
For perpendicular lines slopes are negative reciprocal: -3/2

How you write equation: y - y1 = m(x-x1)
Parallel: y - -1 = 2/3(x-4)
y + 1 = (2/3)x -8/3
y = (2/3)x -11/3

Perpendicular:
y - - 1 = (-3/2)(x-4)
y + 1 = (-3/2)x + 6
y = (-3/2)x + 5

Answer 4

2x – 3y = 9 and the point (4, –1),
y = 2x/3 - 9 <-- has slope of 2/3
y = 2x/3 + b
-1= 2*4/3 + b --> b = -11/3
So y = 2x/3 = 11/3 is line || to given line
y = -3x/2 + b
-1 = -3*4/2 + b --> b = 5
So y = -3x/2 +5 is line perpendicular to given line