Quadrilateral RSTU is reflected across the y-axis and translated down 3 units. Which coordinates best represents S’?

Mathematics

1 answer:

Darina [25.2K]4 years ago

5 0

The coordinates for point S¹ is C. (3, 2).

Step-by-step explanation:

Step 1:

If a coordinate is reflected across an axis, the coordinate values remain constant but the symbols i.e. positive or negative varies according to the quadrant on to which they are reflected.

The current quadrilateral is in the second quadrant. If it reflected across the y-axis it will be reflected on to the first quadrant.

The first quadrant has a positive x and y coordinate value.

Step 2:

The coordinate of point S is (-3, 5) in the second quadrant.

If it is reflected onto the first quadrant, the coordinates become (3, 5).

If the point is translated down three units it means the y values become (y-3) while the x value remains the same.

So the coordinate for point S¹ = (3, 5-3) = (3, 2) which is option C.

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Use lagrange multipliers to find the shortest distance, d, from the point (4, 0, −5 to the plane x y z = 1

Varvara68 [4.7K]

I assume there are some plus signs that aren't rendering for some reason, so that the plane should be $x+y+z=1$ .

You're minimizing $d(x,y,z)=\sqrt{(x-4)^2+y^2+(z+5)^2}$ subject to the constraint $f(x,y,z)=x+y+z=1$ . Note that $d(x,y,z)$ and $d(x,y,z)^2$ attain their extrema at the same values of $x,y,z$ , so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.

The Lagrangian is

$L(x,y,z,\lambda)=(x-4)^2+y^2+(z+5)^2+\lambda(x+y+z-1)$

Take your partial derivatives and set them equal to 0:

$\begin{cases}\dfrac{\partial L}{\partial x}=2(x-4)+\lambda=0\\\\\dfrac{\partial L}{\partial y}=2y+\lambda=0\\\\\dfrac{\partial L}{\partial z}=2(z+5)+\lambda=0\\\\\dfrac{\partial L}{\partial\lambda}=x+y+z-1=0\end{cases}\implies\begin{cases}2x+\lambda=8\\2y+\lambda=0\\2z+\lambda=-10\\x+y+z=1\end{cases}$

Adding the first three equations together yields

$2x+2y+2z+3\lambda=2(x+y+z)+3\lambda=2+3\lambda=-2\implies \lambda=-\dfrac43$

and plugging this into the first three equations, you find a critical point at $(x,y,z)=\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)$ .

The squared distance is then $d\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)^2=\dfrac43$ , which means the shortest distance must be $\sqrt{\dfrac43}=\dfrac2{\sqrt3}$ .