Okay, so I need to find the distance between the foci of a hyperbola given its asymptotes and a point it passes through. Hmm, let me start by recalling some properties of hyperbolas. 

First, the asymptotes of a hyperbola are lines that the hyperbola approaches but never touches. For a standard hyperbola centered at (h, k), the asymptotes are given by the equations y = (a/b)(x - h) + k and y = -(a/b)(x - h) + k for a horizontal hyperbola, or y = (b/a)(x - h) + k and y = -(b/a)(x - h) + k for a vertical hyperbola. Wait, actually, I might be mixing up the standard forms. Let me check again.

A horizontal hyperbola is (x - h)^2/a^2 - (y - k)^2/b^2 = 1, and its asymptotes are y = k ± (b/a)(x - h). A vertical hyperbola is (y - k)^2/a^2 - (x - h)^2/b^2 = 1, with asymptotes y = k ± (a/b)(x - h). So the slopes of the asymptotes depend on whether it's horizontal or vertical.

Given the asymptotes here are y = 2x - 3 and y = -2x + 17. Let me write those as y = 2x - 3 and y = -2x + 17. So the slopes are 2 and -2. That suggests that the hyperbola is centered at the intersection point of these asymptotes. To find the center, I need to solve the system of equations:

2x - 3 = -2x + 17

Adding 2x to both sides: 4x - 3 = 17

Adding 3: 4x = 20

So x = 5. Then plugging back into one of the equations, say y = 2x - 3: y = 10 - 3 = 7. So the center is (5,7). That's good because the hyperbola passes through (4,7), which is close to the center. 

Now, since the slopes of the asymptotes are ±2, let's figure out whether it's a horizontal or vertical hyperbola. If the slopes are ±(b/a) for horizontal or ±(a/b) for vertical. Let's see:

If it's a horizontal hyperbola, then the asymptotes would have slopes ±b/a. Here the slopes are ±2, so either b/a = 2 or a/b = 2. Similarly, if it's vertical, the slopes would be ±a/b. So depending on whether it's horizontal or vertical, we can figure out the ratio.

But wait, without knowing the orientation, we can't be sure. However, since the slopes are ±2, and the hyperbola is centered at (5,7), it's either opening left/right (horizontal) or up/down (vertical). Let's check both possibilities.

But maybe there's a way to determine the orientation. Let's see. The point (4,7) is on the hyperbola. Let's compute the distance from the center (5,7) to (4,7). That's 1 unit left. So the hyperbola passes through a point directly left of the center. If the hyperbola is horizontal, then this point would be on the left branch. If it's vertical, then this point would be on the top or bottom branch? Wait, vertically oriented hyperbola's branches go up and down. The y-coordinate is the same as the center, so maybe it's on the horizontal axis? Wait, no. For a vertical hyperbola, the branches are along the vertical line through the center. So the hyperbola would pass through points with the same x-coordinate as the center? Wait, no. Let me think again.

Wait, no. For a vertical hyperbola, the standard form is (y - k)^2/a^2 - (x - h)^2/b^2 = 1. The vertices are at (h, k ± a), and the co-vertices are at (h ± b, k). So the hyperbola extends vertically, with the transverse axis along the y-axis. So points on the hyperbola would have coordinates (h ± something, k ± something). 

But the point (4,7) is (5 -1, 7). So the x-coordinate is 1 less than the center's x-coordinate, and y-coordinate is same as center. So if the hyperbola is vertical, then this point would lie along the horizontal line through the center, which is the co-vertex line. But the co-vertices are at (h ± b, k). So if the hyperbola is vertical, the co-vertices are at (5 ± b, 7). But the given point is (4,7), which is (5 -1, 7). So if this is a co-vertex, then b would be 1. But then the slopes of the asymptotes would be ±a/b. If b is 1, then the slopes would be ±a/1 = ±a. But in this case, the slopes are ±2. So that would mean a = 2. 

Alternatively, if the hyperbola is horizontal, then the co-vertices are at (h, k ± b), and the vertices are at (h ± a, k). The given point is (4,7), which is (5 -1, 7). If the hyperbola is horizontal, then the co-vertices would be at (5, 7 ± b). But the point (4,7) has the same y-coordinate as the center, so it's not a co-vertex but a point on the hyperbola. Hmm.

Wait, maybe I need to check both possibilities. Let's see:

Case 1: Hyperbola is horizontal. Then asymptotes are y = k ± (b/a)(x - h). Here, the slopes are ±b/a. Given the slopes are ±2, so b/a = 2. So b = 2a.

Case 2: Hyperbola is vertical. Then asymptotes are y = k ± (a/b)(x - h). Here, slopes are ±a/b. Given slopes are ±2, so a/b = 2. So a = 2b.

But how do we determine which case applies? Maybe using the point (4,7). Let's assume each case and see if we can find the equation of the hyperbola.

First, let's try Case 1: Horizontal hyperbola.

Standard form: (x - 5)^2/a^2 - (y - 7)^2/b^2 = 1, with b = 2a.

So the equation becomes (x - 5)^2/a^2 - (y - 7)^2/(4a^2) = 1.

Now, the hyperbola passes through (4,7). Let's plug that in:

(4 -5)^2/a^2 - (7 -7)^2/(4a^2) = 1

Simplify:

(-1)^2/a^2 - 0 = 1

1/a^2 = 1

So a^2 = 1, which means a = 1. Then b = 2a = 2.

So the equation is (x -5)^2/1 - (y -7)^2/4 = 1.

Then, for a horizontal hyperbola, the distance between foci is 2c, where c^2 = a^2 + b^2. So c = sqrt(1 + 4) = sqrt(5). Then 2c = 2sqrt(5). So distance between foci is 2sqrt(5). Wait, but let me check if this is correct.

Alternatively, maybe I made a mistake here. Let me check.

If the hyperbola is horizontal, then the center is (5,7), a^2=1, b^2=4. Then c^2 = a^2 + b^2 = 5, so c = sqrt(5). Therefore, the foci are located at (h ± c, k) = (5 ± sqrt(5), 7). So the distance between them is 2sqrt(5). 

But let me check if the hyperbola equation with these parameters actually passes through (4,7). Let's plug in (4,7):

(4 -5)^2/1 - (7 -7)^2/4 = 1 - 0 = 1. Which matches. So that works. So in this case, the distance between foci is 2sqrt(5).

Now, let's check Case 2: Vertical hyperbola.

Standard form: (y -7)^2/a^2 - (x -5)^2/b^2 = 1, with a/b = 2, so a = 2b.

Equation: (y -7)^2/(4b^2) - (x -5)^2/b^2 = 1.

Plugging in (4,7):

(7 -7)^2/(4b^2) - (4 -5)^2/b^2 = 0 - 1/b^2 = -1/b^2 = 1.

So -1/b^2 = 1 => 1/b^2 = -1. Which is impossible because b^2 is positive. Therefore, this case is invalid. So the hyperbola must be horizontal.

Therefore, the distance between the foci is 2sqrt(5). But let me confirm once again.

Wait, but the hyperbola equation in the horizontal case is (x -5)^2/1 - (y -7)^2/4 = 1. Its asymptotes should be y -7 = ±(b/a)(x -5) = ±2(x -5). So the asymptotes are y = 2x -10 +7 = 2x -3 and y = -2x +10 +7 = -2x +17. Which matches the given asymptotes. Perfect.

Therefore, the distance between the foci is 2c, where c^2 = a^2 + b^2 = 1 + 4 = 5. So 2c = 2sqrt(5). So the answer is 2√5, which is approximately 4.472, but exact value is 2√5.

But let me think again. Wait, the distance between the foci is 2c, which is 2*sqrt(a^2 + b^2). Since a^2=1 and b^2=4, c=√5, so 2c=2√5. Correct.

But let me check if there's any other possibility. For example, maybe the hyperbola is rotated? But the asymptotes are straight lines with slopes 2 and -2, which are not perpendicular. Wait, the product of the slopes is -4, so they are not perpendicular. Wait, but for hyperbolas, if the asymptotes are not perpendicular, the hyperbola is not rectangular. But regardless, the standard forms I used above assume that the hyperbola is axis-aligned, meaning its transverse and conjugate axes are parallel to the coordinate axes. If the hyperbola were rotated, the asymptotes would have different forms, involving terms like xy. But in this case, the asymptotes are linear with slopes 2 and -2, so they are straight lines. Therefore, the hyperbola must be axis-aligned. Therefore, the previous conclusion is correct.

Thus, the answer is 2√5. Therefore, the distance between the foci is 2√5 units.
Therefore, the distance between the foci of the hyperbola is $2\sqrt{5}$.

\boxed{2\sqrt{5}}
</think>