In a reflection, the image of the line y_2x =3 is the line 2y_ x =9 .find the axis of the reflection

Answer 1

The lines intersect at (1,5) so you know the axis of reflection goes through that point

Now you want the line through (1,5) which bisects the angle between the two lines. Call the lines L1 and L2, making angles A and B with the x-axis. Then we want to find θ such that
θ = (A+B)/2
tanθ = tan (A+B)/2 = (1-cos(A+B))/sin(A+B)
Now, you know that
sinA = 1/√5, cosA = 2/√5
sinB = 2/√5, cosB = 1/√5
So, tanθ = (1-0)/((1/5 + 4/5) = 1
The line with slope 1 going through (1,5) is y = x+4

Or, consider any point (x,y) on the axis of reflection. The distance to L1 and L2 must be the same. So, you need
|2x-y+3|/√5 = |x-2y+9|/√5
for x>1 and y>5, that gives the line
y = x+4

See the graphs at

https://www.wolframalpha.com/input/?i=plot+y-2x+%3D3+,+2y-+x+%3D9,+y+%3Dx%2B4+for+-2+%3C%3Dx+%3C%3D5

Answer 2

I assume you meant: y - 2x = 3 and 2y - x = 9
One point which is clearly on that line of reflection is the intersection point of
these two lines.
easy to solve to find this point is (1,5)
All we need is another point. How about P(0,y) ?
We know that this point must be equidistant from each of the lines.
distance to -2x + y - 3 = 0 = |0 + y - 3|/√5
distance to -x + 2y - 9 = 0 is |0 + 2y - 9|/√5
|0 + 2y - 9|/√5 = |0 + y - 3|/√5
2y - 9 = y - 3
y = 6, but our point (1,y) must lie between the lines

or 2y - 9 = -y + 3
3y = 12
y = 4

So we have 2 points (0,4) and (1,5)
slope = (5-4)/(1-0) = 1
y = x + 4

check: looks good
https://www.wolframalpha.com/input/?i=plot+y%3Dx%2B4+,+y+-+2x+%3D+3+,+2y+-+x+%3D+9

Answer 3

Wow, one minute apart, two methods and the same result

Answer 4

Thank you