Find the slope between the two points. Problem Solution (4,6) and (5,0) (-3,-4) and (-2,-10)

Rewrite the Cartesian equation y=-3 as a polar equation.

andreev551 [17]

r

sin

θ

=

−

3

Explanation:

Imagine we have a point

P

with Rectangular (also called Cartesian) coordinates

(

x

,

y

)

and Polar coordinates

(

r

,

θ

)

.

The following diagram will help us visualise the situation better:

https://keisan.casio.com/exec/system/1223526375

We can see that a right triangle is formed with sides

x

,

y

and

r

, as well as an angle

θ

.

We have to find the relation between the Cartesian and Polar coordinates, respectively.

By Pythagora's theorem, we get the result

r

2

=

x

2

+

y

2

The only properties we can say about

θ

are its trigonometric functions:

sin

θ

=

y

/

r

⇒

y

=

r

sin

θ

cos

θ

=

x

/

r

⇒

x

=

r

cos

θ

So we have the following relations:

⎧

⎪

⎨

⎪

⎩

r

2

=

x

2

+

y

2

y

=

r

sin

θ

x

=

r

cos

θ

Now, we can see that saying

y

=

−

3

in the Rectangular system is equivalent to say

r

sin

θ

=

−

3

Answer link

Jim G.

May 19, 2018

r

=

−

3

sin

θ

Explanation:

to convert from

cartesian to polar

∙

x

=

r

cos

θ

and

y

=

r

sin

θ

⇒

r

sin

θ

=

−

3

⇒

r

=

−

3

sin

θ

4 0

3 years ago

I need help finding the area

aniked [119]

<h3>Given :</h3>

Base of triangle = 7 yd
Height of triangle = 10 yd

$\\ \\$

Area of triangle

$\\ \\$

We know:-

When base and height of triangle is given we use this formula:

$\bigstar \boxed{ \rm Area \: of \: triangle = \frac{base \times height}{2} }$

$\\ \\$

So:-

$\\ \\$

$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{base \times height}{2} \\$

$\\ \\$

$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 10}{2} \\$

$\\ \\$

$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 5 \times2 }{2} \\$

$\\ \\$

$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 5 \times\cancel2 }{\cancel2} \\$

$\\ \\$

$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 5 \times1 }{1} \\$

$\\ \\$

$\dashrightarrow \sf \: Area \: of \: triangle =7 \times 5$

$\\ \\$

$\dashrightarrow \bf \: Area \: of \: triangle =35 {yd}^{2} \\$

$\\ \\$

$\therefore \underline{\textsf{ \textbf {\: Area \: of \: triangle = \red{35}}} { \red{\bf{yd} }^{ \red2} }}$

$\\ \\$

$\small\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf \small{Formulas\:of\:Areas:-}}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}\end{gathered}$ ]