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# Q. Consider a variable line 'L' which passes through the point of intersection 'P' of...

user8235304 | Student, Grade 11 | Valedictorian

Posted July 27, 2013 at 4:50 PM via web

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Q. Consider a variable line 'L' which passes through the point of intersection 'P' of the lines `3x + 4y -12=0` and `x+2y - 5=0` meeting the co-ordinate axes at points A and B.

Locus of the middle point of the segment AB has the equtaion:_

a) `3x+4y=4xy`

b) `3x+4y=3xy`

c) `4x+3y=4xy`

d) `4x+3y=3xy`

Tagged with math, straight lines

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted July 27, 2013 at 5:35 PM (Answer #1)

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You should solve the system of equations `3x + 4y - 12 = 0` and `x + 2y - 5 = 0` to evaluate the coordinates of intersection point P, such that:

`{(3x + 4y - 12 = 0),(x + 2y - 5 = 0):} => {(3x + 4y = 12),(x + 2y = 5):} => {(3x + 4y = 12),(x = 5 - 2y):}`

Replacing `5 - 2y` for x in the top equation yields:

`3(5 - 2y) + 4y = 12 => 15 - 6y + 4y = 12`

`-2y = 12 - 15 => -2y = -3 => y = 3/2`

`x = 5 - 2*3/2 => x = 2`

Hence, evaluating the coordinates of the point P yields `x = 2, y = 3/2` .

The line that passes through the point P intersects x axis at the point `A (x_A,0)` , hence, you need to evaluate the equation of the line `PA` , such that:

`(x_A - x_P)/(x - x_P) = (y_A - y_P)/(y - y_P)`

`(x_A - 2)/(x - 2) = (0 - 3/2)/(y - 3/2)`

`-3/2(x - 2) = (y - 3/2)(x_A - 2) => -3x/2 + 3 = y*x_A - 2y - (3/2)x_A + 3`

Reducing duplicate members yields:

`-3x/2 + 2y = x_A(y + 3/2) => x_A = (-3x + 4y)/(2y + 3)`

The line that passes through the point P intersects y axis at the point `B (0,y_B)` , hence, you need to evaluate the equation of the line `BA` , such that:

`(x_B - x_P)/(x - x_P) = (y_B - y_P)/(y - y_P)`

`-2/(x - 2) = (y_B - 3/2)/(y - 3/2)`

`(x - 2)(y_B - 3/2) = -2y + 3 => xy_B - 3x/2 - 2y_B + 3 =-2y + 3`

Reducing duplicate members yields:

`xy_B - 3x/2 - 2y_B =-2y => y_B(x - 2) = 3x/2 - 2y`

`y_B = (3x - 4y)/(2x - 4)`

You need to evaluate the coordinates of midpoint of the line AB, such that:

`x_M = (x_A + x_B)/2 => x_M = (-3x + 4y)/(4y + 6)`

`y_M = (y_A + y_B)/2 => y_M = (3x - 4y)/(4x - 8)`

You need to evaluate the equation of the line MP, such that:

`(x_M - x_P)/(x - x_P) = (y_M - y_P)/(y - y_P)`

`((-3x + 4y)/(4y + 6) - 2)/(x - 2) = ((3x - 4y)/(4x - 8) - 3/2)/(y - 3/2)`

`(-3x + 4y - 8y - 12)/((4y + 6)(x - 2)) = (6x - 8y - 12x + 24)/((4x - 8)(2y - 3))`

`(-3x - 4y - 12)(4x - 8)(2y - 3) = (24 - 6x - 8y)(4y + 6)(x - 2)`

`(-3x - 4y - 12)(8xy - 12x - 16y + 24) = (24 - 6x - 8y)(4xy - 8y + 6x - 12)`

`-24x^2y + 36x^2 + 48xy - 72x - 32xy^2 + 48xy + 64y^2 - 96y - 96xy + 144x + 192y - 288 = 96xy - 192y + 144x - 288 - 24x^2y + 48xy - 36x^2 + 72x - 32xy^2 + 64y^2 - 48xy + 96y`

Reducing like terms yields:

`- 72x + 48xy + 96y - 96xy = 96xy + 72x - 48xy - 96y`

`144x + 192y = 192xy`

`12x + 16y = 16xy => 3x + 4y = 4xy`

Hence, evaluating the locus of the midpoint of the segment AB, under the given conditions, yields `3x + 4y = 4xy` , thus, you need to select the answer a).

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