Write equation of the lines that pass through the point (3,5) and are parallel to and perpendicular to 2x-3y=1

Question

Thank you =]

Answer 1

The equation of any line can have the form y = mx + b, where m is slope, and b is y-intercept.

Put the given line in the form y = mx + b:
2x - 3y = 1
2x - 1 = 3y
2/3 x - 1/3 = y

Parallel lines have the same slope.
So your line has the slope 2/3
y = 2/3 x + b
You now need to know the y-intercept for your line. Plug in the given point for x and y
5 = 2/3 (3) + b

and solve for b

5 = 2 + b
3 = b

so your parallel line is y = 2/3 x + 3

Perpendicular lines have slopes that are opposite reciprocals of eachother.
The slope of the given line is 2/3. The opposite reciprocal of 2/3 is -3/2.
So the slope of your perpendicular line is -3/2

y = -3/2 x + b

Now just plug in the given point for x and y, and solve for b

5 = -3/2 (3) + b
5 = -9/2 + b
5 + 9/2 = b
19/2 = b
9 1/2 = b

So your perpendicular line is y = -3/2 x + 9 1/2

Answer 2

the trick to this problem is to find the slope of the parallel line, then find the y intersects for the parallel and perpendicular line.

The slope of the line is in the equation shown. Set up to y(x) = mx+b:

2x - 3y = 1 => 3y = 2x - 1

y(x) = 2/3 x -1

so... at 3,5, you have to move 3 back and you want to find the y intercept (b value) for the lines.

Parallel line:

5 - 2/3*3 = 3 hence y(x) = 2/3 x + 3


Perpendicular Line (this slope is perp to teh parallel line)

m = -3/2

5 - -3/2*3 = 5 + 9/2 = 19/2 = b hence y(x) = -3/2 x + 19/2


hope that helps :-)

Answer 3

use the graph for this. first find values for x and y in the equation 2x-3y=1 and then locate on the graph. i have got two values enough for location, its x = 2, y=0 & x=3, y=-2. then locate (3,5) and draw the perpendiculars and parallels from this point to the line of the equation 2x-3y=1. after getting the values from the parallel and perpendicular lines dervie the equation from the values.

Answer 4

2x - 3y = 1
-3y = 1 - 2x
y = (2/3)x - (1/3)

Slope of parallel line is 2/3
Slope of perpendicular line is -3/2

5 = (2/3)*3 + b
5 = 2 + b
3 = b

5 = (-3/2)*3 + b
5 = (-9/2) + b
19/2 = b

The equation of the parallel line is y = (2/3)x + 3
The equation of the perpendicular line is y = (-3/2)x + (19/2)

Answer 5

2x - 3y = 1
-3y = -2x + 1
y = (2/3)x - (1/3)

Parallel
(3,5), m = (2/3)
5 = (2/3)(3) + b
5 = 2 + b
b = 3

y = (2/3)x + 3

Perpendicular
(3,5), m = (-3/2)
5 = (-3/2)(3) + b
5 = (-9/2) + b
10 = -9 + 2b
19 = 2b
b = (19/2)

y = (-3/2)x + (19/2)

Parallel : y = (2/3)x + 3
Perpendicular : y = (-3/2)x + (19/2)

Answer 6

First, change 2x-3y=1 to slope intercept form.
-3y=1-2x
y=(2/3)x-(1/3)
Thus, the lines will have slopes 2/3 and -3/2.
The equations will be, in point slope form
y-5=(2/3)(x-3) and
y-5=(-3/2)(x-3)