Answer: To solve this problem, we can start by drawing a diagram of the two triangles and the line connecting their hypotenuses. Let's label the height of the first triangle as h1, the width of the second triangle as w2, and the slope of the line as m.
We know that the hypotenuse of a right triangle is always longer than either of its legs, so we can conclude that h1 < w2. We also know that the height of the second triangle is 12 units, so we can label that as h2 = 12.
Using the Pythagorean theorem, we can express the lengths of the hypotenuses in terms of their legs:
h1^2 + 3^2 = L^2, where L is the length of the first hypotenuse
w2^2 + 12^2 = L^2, where L is the length of the second hypotenuse
Setting these two expressions equal to each other, we can solve for h1 in terms of w2:
h1^2 + 3^2 = w2^2 + 12^2
h1^2 = w2^2 + 12^2 - 3^2
h1^2 = w2^2 + 141
h1 = sqrt(w2^2 + 141)
Now, we can use the fact that the hypotenuses lie on the same line to write an equation for that line:
y = mx + b
where b is the y-intercept of the line. Since the hypotenuses intersect the y-axis at the points (0,h1) and (0,12), we know that b = (h1 + 12)/2. Therefore, our equation becomes:
y = mx + (h1 + 12)/2
To find the slope of the line, we can use the fact that the rise (change in y) over run (change in x) is the same for both triangles. We can express this as:
(h1 - 12) / 3 = 12 / w2
Solving for h1 in terms of w2 and simplifying, we get:
h1 = (36/w2) + 12
Substituting this expression for h1 into our equation for the line, we get:
y = mx + ((36/w2) + 24)/2
y = mx + (18/w2) + 12
Comparing this to the general form of the equation for a line, y = mx + b, we see that the slope is simply:
m = 18/w2
Substituting w2 = 3 (since we were given that the width of the second triangle is 3 units), we get:
m = 18/3 = 6
Therefore, the slope of the line is 6.
Explanation: