Anonymous
Anonymous asked in 科學數學 · 6 years ago

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

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  • 6 years ago
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    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

  • 6 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
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