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.
(someone plz help cuz i don't understand)
To prove that the diagonals of a square are perpendicular to each other, we need to show that the product of their slopes is -1.
Let's take the coordinates of the vertices of the square: A(0,0), B(a,0), C(a,a), and D(0,a).
The slope of a line passing through two points (x₁, y₁) and (x₂, y₂) is given by the formula:
slope = (y₂ - y₁) / (x₂ - x₁)
Let's calculate the slopes of the diagonals AC and BD:
1. Slope of AC:
- Take the coordinates of points A(0,0) and C(a,a).
- Using the slope formula, we have:
slope(AC) = (a - 0) / (a - 0) = a/a = 1
2. Slope of BD:
- Take the coordinates of points B(a,0) and D(0,a).
- Using the slope formula, we have:
slope(BD) = (a - 0) / (0 - a) = -a/a = -1
Now, we can see that the slope of AC is 1 and the slope of BD is -1. The diagonals of a square are perpendicular if the product of their slopes is -1.
Product of slopes: slope(AC) * slope(BD) = 1 * (-1) = -1
Since the product of the slopes is -1, we can conclude that the diagonals AC and BD of the square are perpendicular to each other.