Let's complete the coordinate proof step-by-step:
1. Assignment of coordinates:
Since [tex]\( G H J K \)[/tex] is a square, the coordinates of the vertices can be taken as:
- [tex]\( G (0, 0) \)[/tex]
- [tex]\( H (a, 0) \)[/tex]
- [tex]\( J (a, a) \)[/tex]
- [tex]\( K (0, a) \)[/tex]
Therefore, the coordinates of [tex]\( H \)[/tex] are [tex]\( (a, 0) \)[/tex].
2. Slope of [tex]\( \overline{KH} \)[/tex]:
- The coordinates of [tex]\( K \)[/tex] are [tex]\( (0, a) \)[/tex], and the coordinates of [tex]\( H \)[/tex] are [tex]\( (a, 0) \)[/tex].
- The slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\][/tex]
- Thus, the slope of [tex]\( \overline{KH} \)[/tex] is:
[tex]\[
m_{KH} = \frac{0 - a}{a - 0} = \frac{-a}{a} = -1
\][/tex]
3. Slope of [tex]\( \overline{GJ} \)[/tex]:
- The coordinates of [tex]\( G \)[/tex] are [tex]\( (0, 0) \)[/tex], and the coordinates of [tex]\( J \)[/tex] are [tex]\( (a, a) \)[/tex].
- The slope of [tex]\( \overline{GJ} \)[/tex] is:
[tex]\[
m_{GJ} = \frac{a - 0}{a - 0} = \frac{a}{a} = 1
\][/tex]
4. Product of the slopes:
- The product of the slopes of [tex]\( \overline{KH} \)[/tex] and [tex]\( \overline{GJ} \)[/tex] is:
[tex]\[
m_{KH} \times m_{GJ} = -1 \times 1 = -1
\][/tex]
Since the product of the slopes of two lines is [tex]\(-1\)[/tex], the lines are perpendicular to each other.
5. Conclusion:
Therefore, [tex]\( \overline{KH} \)[/tex] is perpendicular to [tex]\( \overline{GJ} \)[/tex].
Using this information, we can fill in the given blanks:
- Since [tex]\( GHJK \)[/tex] is a square, the coordinates of [tex]\( H \)[/tex] are [tex]\((a, \ldots . . . . . . . . . . .\)[/tex].
- The coordinates of [tex]\( H \)[/tex] are [tex]\((a, 0)\)[/tex].
- The slope of [tex]\( \overline{KH} \)[/tex] is [tex]\(\ \ -1\)[/tex].
- The slope of [tex]\(\ \ \overline{GJ} \ \ \)[/tex] is [tex]\( \ \ 1\)[/tex].
- The product of the slopes of the diagonals is [tex]\( -1 \)[/tex].
- Therefore, [tex]\( \overline{KH} \)[/tex] is perpendicular to [tex]\( \overline{GJ} \)[/tex].
So the completed proof is:
1. Since [tex]\( GHJK \)[/tex] is a square, the coordinates of [tex]\( H \)[/tex] are [tex]\((a, 0)\)[/tex].
2. The slope of [tex]\( \overline{KH} \)[/tex] is [tex]\(-1\)[/tex].
3. The slope of [tex]\( \overline{GJ} \)[/tex] is [tex]\(1\)[/tex].
4. The product of the slopes of the diagonals is [tex]\(-1\)[/tex].
5. Therefore, [tex]\( \overline{KH} \)[/tex] is perpendicular to [tex]\( \overline{GJ} \)[/tex].
