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3 years ago

Three vertices of parallelogram JKLM are J(1, 4), K(5, 3), and L(6,−3). Find the coordinates of vertex M.

Mathematics

1 answer:

Andrej [43]3 years ago

5 0

Answer:

coordinates of vertex M is (x, y) = (2, -2)

Step-by-step explanation:

Since JKLM is a parallelogram, this implies that JK parallel to LM and KL parallel to JM. This means that

Slope of JK = slope of LM

$\frac{3-4}{5-1} =\frac{-3-y}{6-x} \\y=-\frac{1}{4}x-\frac{3}{2} ....(i)$

And

Slope of KL = slope of JM

$\frac{3-\left(-3\right)}{5-6}=\frac{4-y}{1-x}\\y=10-6x...(ii)$

From equation (i) and (ii) we get

$-\frac{1}{4}x-\frac{3}{2} =10-6x$

$-\frac{23x}{4}=-\frac{23}{2}$

$-23x=-46$

$x=2$

Put the value of x in equation (ii) we get

$y=10-6(2)\\y=-2$

So, the coordinates of vertex M is (x, y) = (2, -2).

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Read 2 more answers

Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)

babunello [35]

$\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k$

We want to find $f(x,y,z)$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=x^2+y$

$\dfrac{\partial f}{\partial y}=y^2+x$

$\dfrac{\partial f}{\partial z}=ze^z$

Integrating both sides of the latter equation with respect to $z$ tells us

$f(x,y,z)=e^z(z-1)+g(x,y)$

and differentiating with respect to $x$ gives

$x^2+y=\dfrac{\partial g}{\partial x}$

Integrating both sides with respect to $x$ gives

$g(x,y)=\dfrac{x^3}3+xy+h(y)$

Then

$f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)$

and differentiating both sides with respect to $y$ gives

$y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C$

So the scalar potential function is

$\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}$

By the fundamental theorem of calculus, the work done by $\vec F$ along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it $L$ ) in part (a) is

$\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}$

and $\vec F$ does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them $L_1$ and $L_2$ ) of the given path. Using the fundamental theorem makes this trivial: