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3 years ago

11

Amara wants to draw isosceles right triangles on a coordinate plane that have two vertices at (-2,4) and (5,4). If the coordinat

es of the third vertex are both integers, how many different isosceles right triangles can she draw?
Fill in the banks to explain your reasoning.
For the right triangle to be isosceles and have both coordinates of the third
vertex be integers, a side from one of the two given vertices to the third vertex
must have a length of _____ units and be parallel to the ___-axis.
The y-coordinate for the third vertex must be _____ or _____.

Amara can draw ____ isosceles right triangle(s).

1 answer:

Veseljchak [2.6K]3 years ago

8 0

Answer:

The side must have a length of <u>7</u> units

and be parallel to the <u>y</u>-axis

The y-coordinate for the third vertex must be <u>-3</u> or <u>11</u>

He can draw <u>4</u> isosceles right triangles

Step-by-step explanation:

∵ The side joining (-2 , 4) and (5 , 4) is horizontal (same y-coordinates)

∴ Its length = 5 - 2 = 7

∴ The side from the one of the two given vertices to the third

vertex must be vertical so it's parallel to the y-axis

∴ The x-coordinate will be -2 or 5

∵ The distance between the y-coordinates of the third vertex

and one of the given point is 7

∴ 4 - y = 7 ⇒ y = 4 - 7 = -3

∴ y - 4 = 7 ⇒ y = 4 + 7 = 11

∴ The y-coordinate of the third vertex must be -3 or 11

∴ The third vertex is (-2 , -3) , (-2 , 11) , (5 , -3) , (5 , 11)

∴ Amara can draw 4 isosceles right triangles

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Let the $x$ -coordinate of $P$ be $t$ . For $P\!$ to be on the graph of the function $y = \sqrt{x}$ , the $y$ -coordinate of $\! P$ would need to be $\sqrt{t}$ . Therefore, the coordinate of $P \!$ would be $\left(t,\, \sqrt{t}\right)$ .

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$\begin{aligned} & d\left(\left(t,\, \sqrt{t}\right),\, (9,\, 0)\right) \\ &= \sqrt{(t - 9)^2 +\left(\sqrt{t}\right)^{2}} \\ &= \sqrt{t^2 - 18\, t + 81 + t} \\ &= \sqrt{t^2 - 17 \, t + 81}\end{aligned}$ .

The goal is to find the a $t$ that minimizes this distance. However, $\sqrt{t^2 - 17 \, t + 81}$ is non-negative for all real $t\!$ . Hence, the $\! t$ that minimizes the square of this expression, $\left(t^2 - 17 \, t + 81\right)$ , would also minimize $\sqrt{t^2 - 17 \, t + 81}\!$ .

Differentiate $\left(t^2 - 17 \, t + 81\right)$ with respect to $t$ :

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$\displaystyle \frac{d^{2}}{dt^{2}}\left[t^2 - 17 \, t + 81\right] = 2$ .

Set the first derivative, $(2\, t - 17)$ , to $0$ and solve for $t$ :

$2\, t - 17 = 0$ .

$\displaystyle t = \frac{17}{2}$ .

Notice that the second derivative is greater than $0$ for this $t$ . Hence, $\displaystyle t = \frac{17}{2}$ would indeed minimize $\left(t^2 - 17 \, t + 81\right)$ . This $t\!$ value would also minimize $\sqrt{t^2 - 17 \, t + 81}\!$ , the distance between $P$ $\left(t,\, \sqrt{t}\right)$ and $(9,\, 0)$ .

Therefore, the point $P$ would be closest to $(9,\, 0)$ when the $x$ -coordinate of $P\!$ is $\displaystyle \frac{17}{2}$ .

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Answer:

$(14+12)\times (8+12) > (14\times 12)+(8\times 12)$

Step-by-step explanation:

First let us evaluate the value of both the expressions.

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a) Hence, Larry's claim that both the expressions are equivalent is wrong.

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$(14+12)\times (8+12) > (14\times 12)+(8\times 12)$

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