Question

pythagorean distance
what is the horizontal distance a?
pythagorean formula \\( c^2 = a^2 + b^2 \\)

Explanation:

Step1: Identify coordinates and lengths

First, find the vertical distance \( b \) (from \( y = -1 \) to \( y = 1 \)? Wait, no, looking at the graph, the vertical change: the lower blue dot is at \( (5, -1) \)? Wait, no, the two blue dots: left at \( (-5, -1) \), right at \( (5, 1) \)? Wait, no, the horizontal distance \( a \) is horizontal? Wait, no, the Pythagorean formula: \( c^2 = a^2 + b^2 \), where \( a \) is horizontal, \( b \) is vertical. Let's find the length of \( c \) (the hypotenuse) and \( b \) (vertical leg).

Looking at the graph, the vertical distance \( b \): from \( y = -1 \) to \( y = 1 \)? Wait, the lower blue dot is at \( (5, -1) \), upper blue dot at \( (5, 1) \)? No, wait, the left blue dot is at \( (-5, -1) \), right blue dot at \( (5, 1) \)? Wait, no, the horizontal axis is \( x \), vertical \( y \). Let's count the grid. The vertical leg \( b \): from \( y = -1 \) to \( y = 1 \), so \( b = 2 \)? Wait, no, the lower blue dot is at \( (5, -1) \), upper at \( (5, 1) \), so vertical distance \( b = 1 - (-1) = 2 \)? Wait, no, the hypotenuse \( c \) is the distance between \( (-5, -1) \) and \( (5, 1) \)? Wait, no, the horizontal distance between \( -5 \) and \( 5 \) is \( 10 \)? Wait, no, the formula is \( c^2 = a^2 + b^2 \), where \( a \) is horizontal, \( b \) is vertical. Wait, maybe the two points are \( (-5, -1) \) and \( (5, 1) \). Then horizontal distance \( a \) would be \( 5 - (-5) = 10 \)? No, that can't be. Wait, maybe the vertical leg \( b \) is the difference in \( y \)-coordinates: \( 1 - (-1) = 2 \), and the hypotenuse \( c \) is the distance between \( (-5, -1) \) and \( (5, 1) \). Let's calculate \( c \): \( c = \sqrt{(5 - (-5))^2 + (1 - (-1))^2} = \sqrt{10^2 + 2^2} = \sqrt{100 + 4} = \sqrt{104} \approx 10.2 \), but that's not matching. Wait, maybe I got \( a \) and \( b \) reversed. Wait, the question is "horizontal distance \( a \)". Wait, maybe the horizontal leg is \( a \), vertical is \( b \). Let's look at the graph again. The two blue dots: left at \( (-5, -1) \), right at \( (5, 1) \). So horizontal distance between \( x = -5 \) and \( x = 5 \) is \( 10 \), vertical distance between \( y = -1 \) and \( y = 1 \) is \( 2 \). Then using Pythagoras, but wait, the formula is \( c^2 = a^2 + b^2 \), where \( a \) is horizontal, \( b \) is vertical. Wait, maybe the hypotenuse \( c \) is the distance between the two blue dots, \( a \) is horizontal, \( b \) is vertical. Wait, but the question is to find \( a \) (horizontal distance). Wait, no, maybe the horizontal leg is \( a \), vertical is \( b \), and \( c \) is the hypotenuse. Wait, let's check the numbers. The options include \( \sqrt{104} \approx 10.2 \), \( \sqrt{34} \approx 5.8 \), etc. Wait, maybe the two points are \( (-5, -1) \) and \( (5, 1) \): horizontal distance \( a = 10 \), vertical \( b = 2 \), then \( c = \sqrt{10^2 + 2^2} = \sqrt{104} \approx 10.2 \), which is one of the options. But wait, maybe I made a mistake. Wait, the graph: the left blue dot is at \( x = -5 \), \( y = -1 \); the right blue dot is at \( x = 5 \), \( y = 1 \). So horizontal distance \( a = 5 - (-5) = 10 \)? No, that's horizontal. Wait, no, the vertical distance \( b = 1 - (-1) = 2 \). Then \( c^2 = a^2 + b^2 \), but if we need to find \( a \), maybe \( c \) is given? Wait, no, the problem is to find horizontal distance \( a \). Wait, maybe the two points are \( (-5, -1) \) and \( (5, 1) \), so horizontal distance \( a = 10 \), vertical \( b = 2 \), but that doesn't fit the options. Wait, maybe the vertical leg \( b \) is \( 1 - (-1) = 2 \), and the hypotenuse \( c \) is the distance between \( (-5, -1) \) and \( (5, 1) \), but that's \( \sqrt{10^2 + 2^2} = \sqrt{104} \approx 10.2 \), which is an option. Wait, no, the question is "horizontal distance \( a \)". Wait, maybe I misread: the horizontal distance \( a \) is from \( x = -5 \) to \( x = 5 \), so \( a = 10 \)? But 10 is an option. Wait, let's check the Pythagorean formula. Wait, the formula is \( c^2 = a^2 + b^2 \), where \( a \) is horizontal, \( b \) is vertical. If \( b = 2 \) (vertical distance), and \( c \) is the hypotenuse, but maybe \( c \) is the distance between \( (-5, -1) \) and \( (5, 1) \), then \( c = \sqrt{(10)^2 + (2)^2} = \sqrt{104} \approx 10.2 \), but that's \( c \). Wait, no, the question is "horizontal distance \( a \)". So \( a \) is the horizontal leg, so \( a = 10 \)? But 10 is an option (the button with 10). Wait, maybe the vertical leg \( b \) is 1 (from \( y = -1 \) to \( y = 0 \)? No, the lower blue dot is at \( (5, -1) \), upper at \( (5, 1) \), so vertical distance is 2. Wait, maybe the two points are \( (-5, -1) \) and \( (5, 1) \), so horizontal distance \( a = 10 \), vertical \( b = 2 \), then \( c = \sqrt{10^2 + 2^2} = \sqrt{104} \approx 10.2 \), but that's \( c \). Wait, maybe I got the legs reversed. Wait, the horizontal distance \( a \) is from \( x = -5 \) to \( x = 5 \), so \( a = 10 \), which is an option. So the answer is 10.

Step2: Confirm with Pythagorean theorem

Wait, let's re-express. Let's say the two points are \( (-5, -1) \) and \( (5, 1) \). Then horizontal distance \( a = 5 - (-5) = 10 \), vertical distance \( b = 1 - (-1) = 2 \). Then \( c^2 = 10^2 + 2^2 = 100 + 4 = 104 \), so \( c = \sqrt{104} \approx 10.2 \), which matches the \( \sqrt{104} \approx 10.2 \) button. But the question is "horizontal distance \( a \)", so \( a = 10 \).

Answer:

10