Question

想搬到沒有數學的地方

急！想請問第5題如果需要用旁邊的三角形解的話，應該怎麼列式和計算呢？@@

=> 3 = = 2₁ t = 6
5 ± √₁3
3. 設L:(5-3m)x+(m-1)y=10,L:(2m-1)x-3y = 19,m 為實數,求
(1)L₁//L2 › El m = ___ 8 √ 2 · (2) 4₁ L₂
(Hint:兩直線平行,則斜率;兩直線垂直,則斜率 mm xH= = -1)
m =
2-3 m
M₁ =
=> (m₁-8) (m-²) = 0
(1) -443m
>M-1
-543m 2m-1
"IX
J => -15 tim² = 2-3m4|
4.若直線2x+ay-1=0和直線bx-3y+c=0垂直,則3a-2b+5= 5
M-1
m三一六号
30-²6 +5 = 5
- 1/2 x ² = -1
=-|
=> 2b = 30-
- 3X-Y-1=0
- - X - Y + 1 = 0
=> -5 +3m
M-I
▶ 5. △ABC 中,已知 (2,-1)、B(6,-3) 及垂心H(3,-2),則另一頂點C的坐標為
(Hint:垂心為高的交點,請先畫出示意圖幫助自己思考)
' => Y + 1 = - | { X~2)
M₁5TH = = = }}
= - X - Y + 1 = 0
=> 7+3 = = = (x-6)
= - = X - Y - 1 = 0
MAH = -1
1
3
>m-1.
=> -3m+3 = fm² - 13m45
= 6m² - 10m + 2 = 0
-X-JY-3 = 0
- X - Y + 1 = 0
2Y = 4
Y=2
--
(32)-1)
fc (X,Y)
VH13,-2)
13(6-3)

Answer 1

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qn

CH⟂AB 可以先找它的斜率
AB斜率為-1/2
所以CH斜率為2
得到CH直線方程式 y=2x-8
假設C點座標(t, 2t-8)
AC⟂BH (或用BC⟂AH也可以)
AC斜率=(2t-7)/(t-2)
BH斜率=-1/3
(2t-7)/(t-2) × (-1/3) = -1
2t-7=3t-6
t=-1
所以C點座標為(-1,-10)

想搬到沒有數學的地方

太好了ㅠㅠ 謝謝你願意教我~！