## anonymous one year ago Need help, part and c http://prntscr.com/8cwxa2

1. anonymous

@Michele_Laino

2. Michele_Laino

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: $\Large \begin{gathered} A = - \frac{{B{C^2}}}{2} + \frac{{\pi {r^2}}}{2} = - \frac{{B{C^2}}}{2} + \frac{{\pi B{C^2}}}{4} = \hfill \\ \hfill \\ = \frac{{B{C^2}}}{2}\left( {\frac{\pi }{2} - 1} \right) \hfill \\ \end{gathered}$

3. Michele_Laino

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

4. anonymous

14cm^2

5. anonymous

part c also.

6. Michele_Laino

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?

7. anonymous

@Michele_Laino

8. anonymous

idk...

9. Michele_Laino

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

10. anonymous

x and y ???

11. Michele_Laino

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

12. anonymous

1 min plz..

13. Michele_Laino

ok!

14. anonymous

12y - 3x = 60

15. anonymous

@Michele_Laino

16. Michele_Laino

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

17. anonymous

How?

18. Michele_Laino

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

19. Michele_Laino

into*

20. anonymous

lemme solve plz .. 1 min..

21. Michele_Laino

ok!

22. anonymous

4y - 3x = 12

23. anonymous

what next?

24. Michele_Laino

please wait there is a sign error

25. Michele_Laino

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

26. Michele_Laino

anyway: the requested distance, is: $\Large \begin{gathered} d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} = \hfill \\ \hfill \\ = \sqrt {{{\left( {8 - \left( { - 4} \right)} \right)}^2} + {{\left( { - 3 - 6} \right)}^2}} = \hfill \\ \hfill \\ = \sqrt {{{12}^2} + {9^2}} = ...? \hfill \\ \end{gathered}$

27. anonymous

15 units?

28. Michele_Laino

correct!

29. anonymous

what next?

30. Michele_Laino

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

31. anonymous

3

32. Michele_Laino

yes! it is the point (0,3)

33. anonymous

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

34. Michele_Laino

it is a standard formula

35. anonymous

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

36. Michele_Laino

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

37. Michele_Laino

where m is the slope of our line

38. anonymous

okay.. what next?

39. Michele_Laino

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

40. anonymous

now we equate them , right?

41. Michele_Laino

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

42. Michele_Laino

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

43. anonymous

next?

44. Michele_Laino

we have finished, since we got the standard formula

45. anonymous

thanks.

46. Michele_Laino

here is the situation of your exercise: |dw:1441473727673:dw|

47. Michele_Laino

next we have to compute the subsequent distances: $\Large \begin{gathered} d(A,C) = \sqrt {{{\left( { - 4 - 0} \right)}^2} + {{\left( {6 - 3} \right)}^2}} = \sqrt {{4^2} + {3^2}} = ...? \hfill \\ \hfill \\ d(B,C) = \sqrt {{{\left( {8 - 0} \right)}^2} + {{\left( { - 3 - 3} \right)}^2}} = \sqrt {{8^2} + {6^2}} = ...? \hfill \\ \end{gathered}$

48. Michele_Laino

$\large \begin{gathered} d(A,C) = \sqrt {{{\left( { - 4 - 0} \right)}^2} + {{\left( {6 - 3} \right)}^2}} = \sqrt {{4^2} + {3^2}} = ...? \hfill \\ \hfill \\ d(B,C) = \sqrt {{{\left( {8 - 0} \right)}^2} + {{\left( { - 3 - 3} \right)}^2}} = \sqrt {{8^2} + {6^2}} = ...? \hfill \\ \end{gathered}$

49. Michele_Laino

the requested ratio, part (i), is given by the subsequent expression: $\Large r = \frac{{d(A,C)}}{{d(B,C)}}$ or by the subsequent ratio: $\Large {r_1} = \frac{{d(B,C)}}{{d(A,C)}} = \frac{1}{r}$