To demonstrate that the diagonals of the square [tex]\(JKLM\)[/tex] are perpendicular, we need to show that the slopes of the diagonals are negative reciprocals of each other. Here’s a step-by-step explanation:
1. Define the points of the square:
- [tex]\(J = (0, 0)\)[/tex]
- [tex]\(K = (4, 0)\)[/tex]
- [tex]\(L = (4, 4)\)[/tex]
- [tex]\(M = (0, 4)\)[/tex]
2. Identify the diagonals:
- Diagonal [tex]\(JL\)[/tex] connects points [tex]\(J\)[/tex] and [tex]\(L\)[/tex].
- Diagonal [tex]\(KM\)[/tex] connects points [tex]\(K\)[/tex] and [tex]\(M\)[/tex].
3. Calculate the slope of diagonal [tex]\(JL\)[/tex]:
- The formula for the slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[
\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}
\][/tex]
- For diagonal [tex]\(JL\)[/tex]:
[tex]\[
\text{slope of } JL = \frac{4 - 0}{4 - 0} = \frac{4}{4} = 1
\][/tex]
4. Calculate the slope of diagonal [tex]\(KM\)[/tex]:
- For diagonal [tex]\(KM\)[/tex]:
[tex]\[
\text{slope of } KM = \frac{4 - 0}{0 - 4} = \frac{4}{-4} = -1
\][/tex]
5. Verify that the slopes are negative reciprocals:
- The product of the slopes of the diagonals should be [tex]\(-1\)[/tex] for them to be perpendicular.
[tex]\[
\text{slope of } JL \times \text{slope of } KM = 1 \times (-1) = -1
\][/tex]
- Therefore, the slopes are negative reciprocals.
Since the product of the slopes [tex]\(1\)[/tex] and [tex]\(-1\)[/tex] is [tex]\(-1\)[/tex], this confirms that the diagonals [tex]\(JL\)[/tex] and [tex]\(KM\)[/tex] are perpendicular. This shows that the diagonals of square [tex]\(JKLM\)[/tex] are indeed perpendicular.
