Question

Complete the given diagram by dragging expressions to each leg of the triangle. Then, correctly complete the equation to derive the distance, d.
anved
d (x2-x1)
(3/2 - 3/1) ď²
-60
(X,Y)
£
d
-2
(x₂ + x1)
6-
Reset
2-
a
-2-
¦+ (y₁ - y)² =¦
(x, y₂)
(x₂-x₁)²
Next
to

Answer 1

The equation to derive the distance d is $\sqrt{(x2-x1)^2+(y2-y1)^2}$. The lengths of the other legs of the given triangle are (y2 - y1) and (x2 - x1).

What is the formula for calculating the distance between two points?

Consider the two points (x1, y1) and (x2, y2)

The formula used for calculating the distance between the two points is

distance = $\sqrt{(x2-x1)^2+(y2-y1)^2}$

Calculation:

Given that,

The triangle in the graph has vertices (x1, y1), (x2, y2), and (x2, y1)

Since this triangle makes 90°, it is a right-angled triangle.

Hypotenuse = (x1, y1) to (x2, y2), Adjacent = (x1, y1) to (x2,y1), and Opposite = (x2, y1) to (x2, y2).

Consider the length of the hypotenuse = d

So, using the distance formula, the length of the hypotenuse(d) is,

$d = \sqrt{(x2-x1)^2+(y2-y1)^2}$

And the lengths of the other two legs of the given triangle are,

Length of the adjacent side: (x1, y1) to (x2,y1)

= $\sqrt{(x2-x1)^2+(y1-y1)^2}$

= $\sqrt{(x2-x1)^2+0}$

= $(x2-x1)$

Length of the opposite side: (x2, y1) to (x2, y2)

= $\sqrt{(x2-x2)^2+(y2-y1)^2}$

= $\sqrt{0+(y2-y1)^2}$

= $(y2-y1)$

Therefore, the derived distances for the given triangle are:

$d=\sqrt{(x2-x1)^2+(y2-y1)^2}$, (x2 - x1), and (y2 - y1).

Learn more about the distance between two points here:

brainly.com/question/661229

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