請教f.5 maths~~ thx~~~~

用戶200412

2010-01-26 6:28 am

請教f.5 maths~~ thx~~~~

1. if the polar coordinates of the points O ,A and B are( 90 ,0度 ) ,(24 ,240 度) and (7 ,330度 ) respectively ,then the perimeter of 三角形OAB is

A. 25
B .48
C. 56
D.62

2.
in the figure ,the straaight line L1 passes through (3,4) .
if the slope of the straight line L2 is -2/9 ,find c

3.

The straight lines x+5y -2 = 0 and L are perpendicular to each other .if (2 ,-3) is a point lying on L ,which of the following is also a point on L?

A. (-3 ,-4)
B.(-3 ,-2 )
C.(3 ,-8)
D (3,2)

4.
if the coordinates of the centre of a circle are( -1 ,6) and its diameter is 4 ,then the equation of the circle is

A. x^2 + y^2 -2x + 12y +21 =0
B .x^2 + y^2 -2x +12y +33 =0
C. x^2 + y^2 + 2x - 12y +21 =0
D. x^2 + y^2 + 2x - 12y +33 =0

5.
consider the circle x^2 + y^2 -2x +4y -32=0.find the equation of the chord whose mid-point is (-1 ,3)
a. 2x-5y +17 =0
B. 3x-7y +24 =0
C.5x +2y -1= 0
D. 7x +3y -2 =0

6. find the area of the region bounded by the circle x^2 + y^2 -4x +6y -2=0

回答 (1)

用戶9112

2010-01-26 8:18 am

✔ 最佳答案

(1) O = (0,0°), A = (24, 240°), B = (7,330°)
OA = 24, OB = 7
Angle AOB = 330° - 240° = 90°
Then by Pythagora’s theorem AB^2 = 24^2 + 7^2 = 625
AB = 25
Perimeter = 24 + 7 + 25 = 56 (C)
(2) Need the figure in order to calculate.
(3) Slope of x + 5y – 2 = 0 is -1/5
Slope of L is 5
For answer (D), the slope of the line joining (2, -3) and (3,2) is (2 + 3)/(3 – 2) = 5. Hence (D) is the answer
(4) Radius = 4/2 = 2
Equation of circle is (x + 1)^2 + (y – 6)^2 = 2^2
x^2 + y^2 + 2x – 12y + 1 + 36 = 4
x^2 + y^2 + 2x – 12y + 33 = 0 (D)
(5) Centre of the circle is (2/2, -4/2) or (1, -2)
Slope of the line joining the centre and the mid-point is (-2-3)/(1+1) = -5/2
Slope of the chord = 2/5
Equation of the chord is (y – 3)/(x + 1) = 2/5
5y – 15 = 2x + 2
2x – 5y + 17 = 0 (A)
(6) x^2 + y^2 – 4x + 6y – 2 = 0
(x^2 – 4x + 4) + (y^2 + 6y + 9) = 2 + 4 + 9
(x – 2)^2 + (y + 3)^2 = 15
Radius = √15
Area = π(√15)^2 = 15π

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