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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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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