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3 years ago

The coordinates of quadrilateral PQRS are P(–3, 0), Q(0, 4), R(4, 1), and S(1, –3). What best describes the quadrilateral?

Mathematics

1 answer:

denis-greek [22]3 years ago

4 0

Answer:

Square

Step-by-step explanation:

Plot the vertices of the quadrilateral PQRS on the coordinate plane (see attached diagram). The diagram shows that this is a square. Let's prove it.

1. Find all sides lengths:

$PQ=\sqrt{(-3-0)^2+(0-4)^2}=\sqrt{(-3)^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5\\ \\QR=\sqrt{(0-4)^2+(4-1)^2}=\sqrt{(-4)^2+(3)^2}=\sqrt{16+9}=\sqrt{25}=5\\ \\RS=\sqrt{(4-1)^2+(1-(-3))^2}=\sqrt{(3)^2+(4)^2}=\sqrt{9+16}=\sqrt{25}=5\\ \\SP=\sqrt{(1-(-3))^2+(-3-0)^2}=\sqrt{(4)^2+(-3)^2}=\sqrt{16+9}=\sqrt{25}=5$

All sides have the same lengths.

2. Find the slopes of all lines:

$PQ:\ \dfrac{4-0}{0-(-3)}=\dfrac{4}{3}\\ \\QR:\ \dfrac{1-4}{4-0}=-\dfrac{3}{4}\\ \\RS:\ \dfrac{-3-1}{1-4}=\dfrac{4}{3}\\ \\SP:\ \dfrac{0-(-3)}{-3-1}=-\dfrac{3}{4}$

Since the slopes of PQ and RS are the same, lines PQ and RS are parallel. Since the slopes of QR and SP are the same, lines QR and SP are parallel.

The slopes $\frac{4}{3}$ and $-\frac{3}{4}$ have the product of

$-\dfrac{3}{4}\cdot \dfrac{4}{3}=-1,$

then lines are pairwise perpendicular.

This means PQRS is a square.

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A) Findi

Romashka [77]

Answer:

Part A)

$\displaystyle \frac{dy}{dx}=-\frac{2xy+y^2}{x^2+2xy}$

Part B)

$\displaystyle y=-\frac{5}{8}x+\frac{9}{4}$

Step-by-step explanation:

We have the equation:

$\displaystyle x^2y+y^2x=6$

Part A)

We want to find the derivative of our function, dy/dx.

So, we will take the derivative of both sides with respect to <em>x:</em>

<em /> $\displaystyle \frac{d}{dx}\Big[x^2y+y^2x\Big]=\frac{d}{dx}\big[6\big]$ <em />

The derivative of a constant is 0. We can expand the left:

$\displaystyle \frac{d}{dx}\Big[x^2y\Big]+\frac{d}{dx}\Big[y^2x\Big]=0$

Differentiate using the product rule:

$\displaystyle \Big(\frac{d}{dx}\big[x^2\big]y+x^2\frac{d}{dx}\big[y\big]\Big)+\Big(\frac{d}{dx}\big[y^2\big]x+y^2\frac{d}{dx}\big[x\big]\Big)=0$

Implicitly differentiate:

$\displaystyle (2xy+x^2\frac{dy}{dx})+(2y\frac{dy}{dx}x+y^2)=0$

Rearrange:

$\displaystyle \Big(x^2\frac{dy}{dx}+2xy\frac{dy}{dx}\Big)+(2xy+y^2)=0$

Isolate the dy/dx:

$\displaystyle \frac{dy}{dx}(x^2+2xy)=-(2xy+y^2)$

Hence, our derivative is:

$\displaystyle \frac{dy}{dx}=-\frac{2xy+y^2}{x^2+2xy}$

Part B)

We want to find the equation of the tangent line at (2, 1).

So, let's find the slope of the tangent line using the derivative. Substitute:

$\displaystyle \frac{dy}{dx}_{(2,1)}=-\frac{2(2)(1)+(1)^2}{(2)^2+2(2)(1)}$

Evaluate:

$\displaystyle \frac{dy}{dx}_{(2,1)}=-\frac{4+1}{4+4}=-\frac{5}{8}$

Then by the point-slope form:

$y-y_1=m(x-x_1)$

Yields:

$\displaystyle y-1=-\frac{5}{8}(x-2)$

Distribute:

$\displaystyle y-1=-\frac{5}{8}x+\frac{5}{4}$

Hence, our equation is:

$\displaystyle y=-\frac{5}{8}x+\frac{9}{4}$

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