Firstly, find the intersection point of y = 2x + 3 and y = x + 1

2x + 3 = x + 1
2x - x = 1 - 3
x = -2

So, y = -2 + 1
y = -1

The intersection point is (-2,-1)

Next, realise that all three lines will pass through this point.

The reflected line is a mirror image of the original line. They are both symmetrical about the line y = x + 1

The symmetry line y = x + 1 has a gradient of 1. It is diagonal to the x and y axes (makes an angle of 45° to each axis)

The original line has y-intercept 3
(the coordinates of this point are (0,3)

The symmetry line y = x + 1 has y-intercept 1 (the coordinates of this point are (0,1)

These two points are directly or vertically above/below each other. The distance between them is 2 units.

Since the symmetry line y = x + 1 is diagonal to both axes, this means that when the point (0,3) is reflected, the resultant point will be horizontally to the right of (0,1)

The line segment connecting (0,3) and (0,1) is perpendicular to the line segment connecting (0,1) and the reflected point. This line segment is also equal in length to the first.

So the reflected point will be 2 units to the right of (0,1) and has coordinates (2,1)


Now that we have the coordinates of two points on the reflected line (2,1) and (-2,-1), we can find the gradient

m = (-1-1)/(-2-2) = -2/-4 = ½

Now we can use the formula y - y1 = m(x - x1) to get the equation

y -1 = ½(x - 2)
y = 1 = ½x - 1
y = ½x

This is the equation of the reflected line.

Alternative approach :

When reflected, the y-intercept (0,3) and the resultant point will be equidistant from the symmetry line y = x + 1

We also know that the line segment joining these two points will be perpendicular to the symmetry line (which has gradient 1)

Since the product of the gradient of two perpendicular lines = -1,

Gradient of the line segment = -1/1 = -1

Sub m = -1, and (0,3) into y - y1 = m(x - x1)

y - 3 = -1(x - 0)
y = -x + 3
This is the equation of the line segment.

The symmetry line is the perpendicular bisector of the line segment. We find the intersection of the two lines, which is the midpoint of the line segment.

For y = x + 1 and y = -x + 3

-x + 3 = x + 1
2x = 1
x = 1

Sub x = 1, y = 1 + 1 = 2
Or y = -1 + 3 = 2

The midpoint has coordinates (1,2)

Next, let the coordinates of the reflected point be (a,b)

We can now use the midpoint theorem and (0,3) to find the reflected point.

Recall midpoint of two points = ( (x1+x2)/2 , (y1+y2)/2 )


So,

(1,2) = ( (0 + a)/2, (3 + b)/2 )

1 = (0+a)/2
2 = a

And

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

The reflected point is (2,1)


We can now use (2,1) and (-2,-1) to find the gradient of the reflected line

Recall that (-2,-1) was found earlier in the previous comment and is the point all three lines will pass through.

Gradient of reflected line = (y2-y1)/(x2-x1)
= (-1-1)/(-2-2)
= -2/-4
= ½

Sub either point and m = ½ into y - y1 = m(x - x1)

y - 1 = ½(x - 2)
y - 1 = ½x - 1
y = ½x