hiufung asked in 科學及數學數學 · 1 decade ago

A math-straight lines 3條(15分)

1.Two lines L1: x+y-5=0 and L2: 2x-3y=0 intersect at a point A. Find the equations of the two lines passing through A whose distances from the origin are equal to 2.

2.A family of straight lines is given by the equation y-3+k(x-y+1)=0, where k is real.

(a)Find the equation of a line L1 in the family whose x-intercept is 5.

(b)Find the equation of a line L2 in the family which is parallel to the x-axis.

(c)Find the acute angle between L1 and L2.

3.L is the line y=2x+3.

(a) A line with a slope m makes an angle of 45 degree with L. Find the value(s) of m.

(b) A family of straight lines is given by the equation 2x-3y+2+k(x-y-1)=0 , where k is real.Using (a), find the equation of the line in the family with positive slope whick makes an angle of 45 degree with L.

要做得詳細d ... thx..

1 Answer

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  • 1 decade ago
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    1.

    The equation of the line passing through A is:

    x + y - 5 + k(2x - 3y) = 0

    (1 + 2k)x + (1 - 3k)y - 5 = 0

    Distance from the origin (0, 0) to the line:

    [(1 + 2k)(0) + (1 - 3k)(0) - 5]/√[(1 + 2k)2 + (1 - 3k)2] = 2

    -5/√[1 + 4k + 4k2 + 1 - 6k + 9k2] = 2

    -5/√[2 - 2k + 13k2] = 2

    2√[2 - 2k + 13k2] = -5

    4(2 - 2k + 13k2) = 25

    52k2 - 8k + 8 = 25

    52k2 - 8k - 17 = 0

    (26k - 17)(2k + 1) = 0

    k = 17/26 or k = -1/2

    When k = 17/26:

    [1 + 2(17/26)]x + [1 - 3(17/26)]y - 5 = 0

    (60/26)x - (25/26)y - 5 = 0

    (12/26)x - (5/26)y - 1 = 0

    12x - 5y - 26 = 0

    When k = -1/2:

    [1 + 2(-1/2)]x + [1 - 3(-1/2)]y - 5 = 0

    (5/2)y - 5 = 0

    y - 2 = 0

    Ans: 12x - 5y - 26 = 0 ooroy - 2 = 0

    2.

    (a)

    The equation of the line: y - 3 + k(x - y + 1) = 0

    kx + (1 - k)y + (k - 3) = 0

    When y = 0, x = -(k - 3)/k

    x-intercept = -(k - 3)/k = 5

    -k + 3 = 5k

    6k = 3

    k = 1/2

    L1: (1/2)x + [1 - (1/2)]y + [(1/2) - 3] = 0

    L1: (1/2)x + (1/2)y + (-5/2) = 0

    L1: x + y - 5 = 0

    (b)

    The equation of the line: y - 3 + k(x - y + 1) = 0

    kx + (1 - k)y + (k - 3) = 0

    When the line // x-axis:

    k = 0

    L2: (0)x + [1 - (0)]y + [(0) - 3] = 0

    L2: y - 3 = 0

    (c)

    Slope of L1 = -(1/1) = -1

    Angle between L1 and x-axis = tan-1|-1|= 45o

    Since L2 // x-axis, angle between L1 and L2 = 45o

    3.

    (a)

    Slope of L = 2

    tan45o = (m - 2)/(1 + 2m) ooro tan45o = (2 - m)/(1 + 2m)

    1 = (m - 2)/(1 + 2m) ooro 1 = (2 - m)/(1 + 2m)

    1 + 2m = m - 2 ooro 1 + 2m = 2 - m

    m = -3 ooro 3m = 1

    m = -3 ooro m = 1/3

    (b)

    The line:

    2x - 3y + 2 + k(x - y - 1) = 0

    (2 + k)x - (3 + k)y + (2 - k) = 0

    Slope m = (2 + k)/(3 + k)

    m = 1/3 (positive):

    (2 + k)/(3 + k) = 1/3

    3(2 + k) = 3 + k

    6 + 3k = 3 + k

    2k = -3

    k = -3/2

    The line is:

    [2 + (-3/2)]x - [3 + (-3/2)]y + [2 - (-3/2)] = 0

    (1/2)x - (3/2)y + (7/2) = 0

    x - 3y + 7 = 0

    =

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