F4 maths(1)

1.己知直線L1: 7x-3y+13=0 與直線 L2: 3x-2y+7=0 相交於Q點。若一直線L通過Q和原點,求L的方程。


First,find the point Q by L1 and L2


from L1

7x-3y+13=0

7x=3y-13

x=(3y-13)/7 ...... (1)

sub (1) into L2

3((3y-13)/7)-2y+7=0

9y-39-14y+49=0

-5y+10=0

y=2

sub y=5/4 into L1

7x-3(2)+13=0

x=1

Q(1,2)


Then,as 直線L通過Q和原點 (0,0) ,so we can find L by point Q和原點.


slope of L= (2-0)/(1-0) =2

the eqt. of L:

2(x-1)=y-2

2x-2=y-2

2x-y=0

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2.若直線L1: 3x-y+8=0 和L2相交於x軸上且L1垂直L2。


a.求L1的斜率

L1的斜率= -A/B = 3


b.求L2的x截距。

∵直線L1: 3x-y+8=0 和L2相交於x軸上

∴L2的x截距=L1的x軸距
= -C/A
= -8/3

c.求L2的方程。

∵L1垂直L2

∴slope of L2 X slope of L1= -1
slope of L2= -1/3

th eqt. of L2:

-1/3(x-(-8/3))=y-0
x+8/3= -3y
3x+9y+8=0

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3.已知直線L1: 9x-py-6=0 與直線L2: 12x+2y+q=0 的方程,若L1和L2有無限個交點,求p和q的值。

ML1=ML2
9/p= -12/2
18= -12p
p=3/2

L1的y截距=L2的y截距
-6/p= q/2
-p=3q ...... (1)

sub p=3/2 into (1)
-3/2=3q
-3=6q
q=-1/2

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ur ans in Q3 is incorrect as u have the wrong concept on the step of 比較L1和L2

比較L1和L2

-py+9x-6=2y+12x+q

4/3 (-py+9x-6) =2y+12x+q

4/3*-py+12x-8=2y+12x+q

when both side have 12x ,u can to compare.

2013-08-15 18:21:55 補充：
4. 用平方根法解下列各方程。

(x+3)^2-64=0

x^2+6x+9-64=0

(x+11)(x-5)

X= -11 X=5