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# 求方程式解

In the figure,L1:x+2y-7=0 cut the x-axis at A.The x-intercept of another lineL2 is -3. Supposed L'1 Is perpendicular with L2 and they intersect at P.

a)Find the equation of L2.

b)find the coordinates of A and P

c)Find the area of the shaded region.

如果可以 希望可以提供算式 謝謝

### 2 Answers

- 如露亦如電Lv 76 years agoFavorite Answer
L'1 是否為 L1 的筆誤，假設為L1

a)因為 L2 垂直於L1 ，故可假設L2方程式為

L2： 2x-y=k，k為常數

代y=0，x截距為 k /2 = -3 ==> k= -6

L2方程式為2x-y= -6 =>2x-y+6=0

b)L1:x+2y-7=0=>y=0時，x=7 ==>A(7,0)

{x+2y-7=0 ----(1)

{2x-y+6=0 ----(2)

解(1)(2)得X=-1,y=4 ===>P(-1,4)

c)無圖，假設是L1,L2與X軸圍成的陰影面積

三角形=>底長=7-(-3)=10，高=4(P點)

陰影面積= 1/2* 10* 4 =20

- Paul LiaoLv 76 years ago
In the figure,L1:x+2y-7=0 cut the x-axis at A.The x-intercept of another lineL2 is -3. Supposed L'1 Is perpendicular with L2 and they intersect at P.

a)Find the equation of L2.

設 L2: 2x-y+a=0 (與 L1 垂直)

L2 x軸截距為 - 3, 2x+a=0 --> a=6

L2 : 2x-y+6=0 ...Ans

b)find the coordinates of A and P

x-7=0 ---> x=7, A(7,0) ...Ans

解 x+2y-7=0 ...(1)

2x-y+6=0 ...(2)

(1)+2*(2) 得 5x+5=0 --->x= - 1 --->y=4 --->P( - 1, 4)

c)Find the area of the shaded region.

沒有圖形無法求

Source(s): Paul