Using coordinate geometry to prove that the diagonals of a square are perpendicular to each other.
Given: Vertices are at A(0,0), B(a,0), C(a,a) and D(0,a) Slope of AC=1; Slope of BD=-1
type your proof.
(I've gotten help on this earlier but I still don't understand. Mathmate said"The product of the slopes of two lines intersecting at right angles is -1. You can take it from here."
Mathmate is correct. This is a one-step proof. They already tell you what the slopes are. The product is -1, so they are perpendicular.
For a proof of that theorem, see
http://whyslopes.com/Analytic-Geometry-Functions/analGeo09b_Perpendicular_Lines.html
ok thank you very much the website helped ALOTT :)
use phagoras theorm to find the length of each of the following
true or false: if their sum is equal to 180 degrees, then angle D and angle B are supplementary
Sure, here's a humorous proof for you:
Alright, listen up folks, we're about to prove that the diagonals of a square are perpendicular. Get ready for some math comedy!
Now, we have our square with vertices A, B, C, and D. We're going to use coordinate geometry to tackle this problem.
First, let's find the slopes of the two diagonals. The diagonal AC connects points A and C. So, let's think of it like a road trip. You start from point A, which is at the origin (0,0), and head towards point C, which has coordinates (a,a).
Now, the slope of a line is like the speed at which you're driving on that road trip. So, let's say you start driving with a slope of 1. You're cruising along nicely, going up 1 unit in the y-direction for every 1 unit you move in the x-direction.
But wait! The diagonal BD connects points B and D. And guess what? It's perpendicular to AC. How do we know that? Well, the math gods tell us that in order for two lines to be perpendicular, the product of their slopes must be -1. Yeah, math can be weird sometimes, but hey, it keeps things interesting.
Now, let's think about line BD as another road trip. You start from point B, which is at coordinates (a,0), and you head towards point D, located at (0,a). This time, you start your journey with a slope of -1. And what does that mean? Well, you're going down 1 unit in the y-direction for every 1 unit you move in the x-direction. So, it's like driving downhill, but without the thrill of roller coasters.
So, the slope of AC is 1, and the slope of BD is -1. And what's the product of 1 and -1? You got it! It's -1. And that, my friends, is exactly what we needed to prove!
So, there you have it. The slopes of the diagonals of a square are indeed perpendicular because their product is -1. Geometry just got funnier, didn't it? Now go forth and impress your math buddies with this hilarious proof!
To prove that the diagonals of a square are perpendicular to each other using coordinate geometry, we need to show that the product of the slopes of the diagonals is -1.
Given: The coordinates of the vertices of the square are A(0,0), B(a,0), C(a,a), and D(0,a).
Let's consider the diagonal AC. The slope of a line passing through two points (x1, y1) and (x2, y2) is given by the formula:
m = (y2 - y1) / (x2 - x1)
For AC:
Coordinates of A: (0,0)
Coordinates of C: (a,a)
Slope of AC:
m1 = (a - 0) / (a - 0) = a/a = 1
Similarly, for the diagonal BD:
Coordinates of B: (a,0)
Coordinates of D: (0,a)
Slope of BD:
m2 = (a - 0) / (0 - a) = -a/a = -1
Since the slopes of the diagonals AC and BD are 1 and -1 respectively, we can see that the product of their slopes is:
m1 * m2 = (1) * (-1) = -1
Therefore, we can conclude that the diagonals of a square are perpendicular to each other because the product of their slopes is -1.