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Find the equations of the two straight lines drawn through the point (0,a) on which the...

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user8235304 | Student, Grade 11 | Valedictorian

Posted June 19, 2013 at 5:51 AM via web

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Find the equations of the two straight lines drawn through the point (0,a) on which the perpendicular let fall from the point (2a,2a) are each of length a. Also prove that the equation of the straight line joining the feet of these perpendiculars is y+2x=5a.

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aruv | High School Teacher | Valedictorian

Posted June 19, 2013 at 7:58 AM (Answer #1)

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Let the slope of the required line which posses through (0,a) be m.

Thus equation of the line be

`y-a=mx`

`mx-y+a=0`              (i)

The perpendicular distance from (2a,2a) to (i) is

`+-a=(mxx2a-2a+a)/sqrt(m^2+1)`

`+-a=(a(2m-1))/sqrt(m^2+1)`

`+-1=(2m-1)/sqrt(m^2+1)`

`+-sqrt(m^2+1)=2m-1`

Squaring both side ,we will have

`m^2+1=4m^2+1-4m`

`3m^2-4m=0`

`m(3m-4)=0`

`m=0 or m=4/3`

Thus equation of the lines are

`y-a=0 or y-a=(4/3)x`

`y=a or 3y-3a=4x`

`y=a or 3y=4x+3a`

Equation of line posses through (2a,2a) and perpendicular to y=a

is

`0(y-2a)=x-2a`

`x-2a=0`         (i)

y= a                    (ii)

(i) and (ii) intersect each other at (2a,a).

Thus foot of perpendicular is (2a,a)

Equation of line posses through (2a,2a) and perpendicular to 3y=4x+3a  is

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

`4y-8a=-3x+6a`

`3x+4y=14a`     (iii)

-4x+3y=3a             (iv)

(iii) and (iv) will intersect each other at ((6a)/5,(13a)/5).Thus foot of perpendicular is `((6a)/5,(13a)/5)`

Equation of line posses through (2a,a) and `((6a)/5,(13a)/5)`  is

`y-a=((13a)/5-a)/((6a)/5-2a)(x-2a)`

`y-a=(8a)/(-4a)(x-2a)`

`y-a=-(2)(x-2a)`

`y-a=-2x+4a`

`2x+y=5a`

`` Hence you have required answer.

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user8235304 | Student , Grade 11 | Valedictorian

Posted June 20, 2013 at 6:14 PM (Reply #1)

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Thanks a lot!!

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