determine the equation of all the lines that are perpendicular to 2x+3y-6=0 nd r 4 units from the origin?

Answer 1

This is an interesting problem. Ernest Maxwell's algebra is wrong. Also, begoner made a mistake in his equation y = (3/2)x + 2 by assuming a y-intercept of 2. That's wrong.

First, let's get this equation in slope-intercept form:

2x + 3y - 6 = 0
3y = -2x + 6
y = (-2/3)x + 2

It has a slope of (-2/3), and the y-intercept is +2. This line goes through the points (0,2) and (3,0).

Now draw a line through the origin, parallel to the original line. This new line has the equation y = (-2/3)x (Eq 1). We want the two points on that line that are 4 units from the origin (one point in the second quadrant and one in the fourth).

Suppose those points are (x,y). Using the distance formula,

x^2 + y^2 = 4^2 = 16 (Eq 2)

From Eq 1, y^2 = (4/9)x^2

Substitute that into Eq 2 to get

x^2 + (4/9)x^2 = 16
(13/9)x^2 = 16
x^2 = (16*9) / 13 = 144/13
x = +/- 12/ sqrt(13) = +/- (12/13) sqrt(13)

And from Eq 1,

y = (-2/3)x = -/+ (8/13) sqrt(13)

The lines we're looking for go through the two points

[(12/13) sqrt(13),-(8/13) sqrt(13)] (fourth quadrant), and

[-(12/13) sqrt(13),(8/13) sqrt(13)] (second quadrant)

The slope of all lines perpendicular to the original line is 2/3. Using the point-slope form, the lines you're looking for are

y +/- (8/13) sqrt(13) = (2/3)[x -/+ (12/13) sqrt(13)]

Multiplying this out, we get

3y +/- (24/13) sqrt(13) = 2x -/+ (24/13) sqrt(13)

and putting it in standard form, it becomes

2x - 3y = +/- (48/13) sqrt(13)

Those are your two lines -- one is positive on the right side and one is negative. These two lines are 4 units from the origin, and they're both perpendicular (slope = 2/3) to the original line.

As a final check, let's do the distance formula on those two points we found:

x^2 + y^2 = [+/- (12/13) sqrt(13)]^2 + [-/+ (8/13) sqrt(13)]^2
x^2 + y^2 = 144*13/169 + 64*13/169
x^2 + y^2 = 144/13 + 64/13 = 208/13 = 16 = 4^2

This proves that the lines are each 4 units from the origin.

So, this is a complete solution to your interesting problem.

Answer 2

You have two equations that must be satisfied: that it's perpendicular to 2x+3y-6=0 and that it's four units away from (0, 0). So:

1. 2x + 3y = 6
3y = -2x + 6
y = (-2/3)x + 2

Now that it's arranged in slope-intercept form, you can find the equation of all lines that are perpendicular to it. That means that the slope must be -1/m, m being the current slope. So the equation of the new line is:

y = (3/2)x + 2

Next you need to find the equation of all points which are 4 units from the origin. That means a circle with a radius of 4 centered at (0, 0), or:

x ^ 2 + y ^ 2 = 16
y ^ 2 = 16 - x ^ 2
y = (16 - x ^ 2) ^ 0.5

Now you can plug in the value of y from the first equation into the second:

(3/2)x + 2 = (16 - x ^ 2) ^ 0.5
(9/4)x ^ 2 + 3x + 4 = 16 - x ^ 2
(13/4)x ^ 2 + 3x - 12 = 0

And then you can just use the quadratic formula to solve it.

Answer 3

[1] Original Equation:
2x + 3y - 6 = 0

[2] Rearranged:
y = (2/3)x + 3

[3] Perpendicular & 4 units away from zero.
1) y = (-3/2)x + 4
2) y = (-3/2)x - 4

Answer 4

Nope,you figure it out

Answer 5

IDK