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Martin wants to use coordinate geometry to prove that the opposite sides of a rectangle are congruent. He places parallelogram [tex]$ABCD$[/tex] in the coordinate plane so that [tex]$A$[/tex] is [tex]$(0,0)$[/tex], [tex]$B$[/tex] is [tex]$(a,0)$[/tex], [tex]$C$[/tex] is [tex]$(a,b)$[/tex], and [tex]$D$[/tex] is [tex]$(0,b)$[/tex].

What formula can he use to determine the distance from point [tex]$D$[/tex] to point [tex]$A$[/tex]?

A. [tex]$\sqrt{(0-0)^2+(b-0)^2}=\sqrt{b^2}=b$[/tex]
B. [tex]$(a-a)^2+(b-0)^2=b^2$[/tex]
C. [tex]$\sqrt{(a-a)^2+(b-0)^2}=\sqrt{b^2}=b$[/tex]
D. [tex]$((-0))^2+(b-0)^2=b^2$[/tex]


Sagot :

To determine the distance between two points in the coordinate plane, Martin can use the distance formula. The distance formula between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:

[tex]\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]

In this problem, point [tex]\(D\)[/tex] has coordinates [tex]\((0, b)\)[/tex] and point [tex]\(A\)[/tex] has coordinates [tex]\((0, 0)\)[/tex].

Let's apply the distance formula step-by-step:

1. Identify the coordinates of points [tex]\(D\)[/tex] and [tex]\(A\)[/tex]:
- Coordinates of [tex]\(D\)[/tex]: [tex]\( (0, b) \)[/tex]
- Coordinates of [tex]\(A\)[/tex]: [tex]\( (0, 0) \)[/tex]

2. Substitute the coordinates into the distance formula:
- [tex]\( x_1 = 0 \)[/tex]
- [tex]\( y_1 = b \)[/tex]
- [tex]\( x_2 = 0 \)[/tex]
- [tex]\( y_2 = 0 \)[/tex]

[tex]\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]

3. Substitute the specific coordinates into the formula:

[tex]\[ \text{Distance} = \sqrt{(0 - 0)^2 + (b - 0)^2} \][/tex]

4. Simplify the expression inside the square root:

[tex]\[ \text{Distance} = \sqrt{0 + b^2} \][/tex]

5. Simplify further:

[tex]\[ \text{Distance} = \sqrt{b^2} \][/tex]

6. Since the square root of [tex]\(b^2\)[/tex] is [tex]\(b\)[/tex] (assuming [tex]\(b \geq 0\)[/tex]):

[tex]\[ \text{Distance} = b \][/tex]

Therefore, the correct formula Martin can use to determine the distance from point [tex]\(D\)[/tex] to point [tex]\(A\)[/tex] is:

[tex]\[ \sqrt{(0-0)^2+(b-0)^2}=\sqrt{b^2}=b \][/tex]

Hence, the correct answer is:

A. [tex]\(\sqrt{(0-0)^2+(b-0)^2}=\sqrt{b^2}=b\)[/tex]