P(0,3) is one point on the Y-axis which is 3 units away from the line 4x-3y=6. Another point 3 units away....?

Question

P(0,3) is one point on the Y-axis which is 3 units away from the line 4x-3y=6. Another point 3 units away is Q(0,K) where K=?

No need to show work, but would appreciate it!!

Answer 1

The distance from the point P(0,3) to the line 4x - 3y = 6 is 3 and can be calculated with the distance formula from a point to a line. First rewrite the equation and set it equal to zero. The proceed with the distance calculation. The distance to the line from point (h,k) is:

d = | 4h - 3k - 6 | / √[4² + (-3)²] = | 4h - 3k - 6 | / √25
d = | 4h - 3k - 6 | / 5

For the point at hand we have:

d = | 4*0 - 3*3 - 6 | / 5 = | -15 | / 5 = 15/5 = 3

The distance from the point Q(0,k) also equals three.

d = | 4*0 - 3k - 6 | / 5 = | ±15 | / 5

- 3k - 6 = 15
- k - 2 = 5
k = -7

The other point is (0,-7).

Answer 2

Draw the situation. The line crosses the y axis when x = 0 which gives y = -2 so at the point (0, -2). You will see that the other point must be below the line. The mirror image point must be (0, -7), five units below the line, as the one given in the question is five units above the line.

Answer 3

Your co-ordinates for P & Q are perfect yet dy/dx for P is two*4-2=2 no longer 0 so then utilising formula (y-y1)=m(x-x1) for P y=2x-8 = tangent for Q y=-2x-4 = tangent then the factor the place those 2 meet capacity the x and y co-ordinates could desire to be equivalent so 2x-8=-2x-4 subsequently 4x=4 x=a million and if x=a million y=2*a million-8 =-6