Question

Need help,
part and c
http://prntscr.com/8cwxa2

Answer 1

@Michele_Laino

Answer 2

question b)
the area of the right triangle, is:
\[\Large \frac{{AC \cdot BC}}{2} = \frac{{B{C^2}}}{2}\]
Using the theorem of Pitagora, we have this:
\[\Large r = \frac{{BC}}{{\sqrt 2 }}\]
where \( r \) is the radius of the half-circumference.
So the requested area is:
\[\Large \begin{gathered}
A = - \frac{{B{C^2}}}{2} + \frac{{\pi {r^2}}}{2} = - \frac{{B{C^2}}}{2} + \frac{{\pi B{C^2}}}{4} = \hfill \\
\hfill \\
= \frac{{B{C^2}}}{2}\left( {\frac{\pi }{2} - 1} \right) \hfill \\
\end{gathered} \]

Answer 3

oops... I have made a typo:
\[\Large \frac{{AB \cdot BC}}{2} = \frac{{B{C^2}}}{2}\]

Answer 4

14cm^2

Answer 5

part c also.

Answer 6

we have to write the equation of your line, in order to do that, we can apply this equation:
\[\Large \frac{{y - {y_1}}}{{{y_2} - {y_1}}} = \frac{{x - {x_1}}}{{{x_2} - {x_1}}}\]
what do you get?

Answer 7

@Michele_Laino

Answer 8

idk...

Answer 9

please try, you have to substitute the coordinates of your points into my formula above:
(x1,y1)=(-4,6), and (x2, y2) = (8,-3)

Answer 10

x and y ???

Answer 11

x, and y are the variables, you have only to susbstitute x1, x2, y1, y2 with the coordinates of your points

Answer 12

1 min plz..

Answer 13

ok!

Answer 14

12y - 3x = 60

Answer 15

@Michele_Laino

Answer 16

I got this:
\[\Large y = - \frac{3}{4}x + 3\]

Answer 17

How?

Answer 18

by substitution int my formula:
\[\Large \frac{{y - 6}}{{ - 3 - 6}} = \frac{{x - \left( { - 4} \right)}}{{8 - \left( { - 4} \right)}}\]

Answer 19

into*

Answer 20

lemme solve plz .. 1 min..

Answer 21

ok!

Answer 22

4y - 3x = 12

Answer 23

what next?

Answer 24

please wait there is a sign error

Answer 25

the slope of your equation is 3/4 which is positive, whereas the slope of the requested line has to be negative

Answer 26

anyway: the requested distance, is:

\[\Large \begin{gathered}
d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} = \hfill \\
\hfill \\
= \sqrt {{{\left( {8 - \left( { - 4} \right)} \right)}^2} + {{\left( { - 3 - 6} \right)}^2}} = \hfill \\
\hfill \\
= \sqrt {{{12}^2} + {9^2}} = ...? \hfill \\
\end{gathered} \]

Answer 27

15 units?

Answer 28

correct!

Answer 29

what next?

Answer 30

the equation of the y-axis is \( \Large x=0 \), so the requested intersection point is given by the solution of this algebraic system:
\[\Large \left\{ \begin{gathered}
y = - \frac{3}{4}x + 3 \hfill \\
\hfill \\
x = 0 \hfill \\
\end{gathered} \right.\]

Answer 31

3

Answer 32

yes! it is the point (0,3)

Answer 33

http://prntscr.com/8cxltt
how did you gte that formula

*get

Answer 34

it is a standard formula

Answer 35

never learnt of it.
Can you give me details about the same??

Answer 36

ok!
the equation of the line which passes at point (x1,y1) is:
\[\Large y - {y_1} = m\left( {x - {x_1}} \right)\]

Answer 37

where m is the slope of our line

Answer 38

okay..
what next?

Answer 39

next I require that line has to pass at point (x2,y2) too, so I can write this:
\[{y_2} - {y_1} = m\left( {{x_2} - {x_1}} \right) \qquad \qquad (*)\]

Answer 40

now we equate them , right?

Answer 41

not exactly, I solve equation (*) for m, so I get:
\[\Large m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\]

Answer 42

then I substitute such expression for m into the first equation:
\[\Large y - {y_1} = \left( {\frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right)\left( {x - {x_1}} \right)\]

Answer 43

next?

Answer 44

we have finished, since we got the standard formula

Answer 45

thanks.

Answer 46

here is the situation of your exercise: