A Line through the variable point `A(1+k;2k)` meets the lines `7x+y-16=0; 5x-y-8=0 and `x-5y+8=0` at B;C;D respectively. Prove that AC;AB and AD are in HP.

Explanation

Let the line passing through point A(k+1,2k) make an angle `theta` with positive x-axis direction.
Then equation of line is
`(x-(k+1))/("cos" theta) = (y-2k)/("sin" theta) =r_(1), r_(2), r_(3),`
`"where " AB =r_(1), AC=r_(2),AD=r_(3)`
`therefore B -= ((k+1) + r_(1) "cos " theta, 2k + r_(1) "sin " theta)`
`C -= ((k+1) + r_(2) "cos " theta, 2k + r_(2) "sin " theta)`
`D -= ((k+1) + r_(3) "cos " theta, 2k + r_(3) "sin " theta)`
Points B, C and D satisfy lines 7x+y-16=0, 5x-y-8=0 and x-5y+8=0, respectively.
So, putting these points in the corresponding lines, we get
`therefore r_(1) = (9(1-k))/(7"cos" theta + "sin" theta), r_(2) = (3(1-k))/(5"cos" theta-"sin" theta)`
`" and " r_(3) = (9(1-k))/(5"sin" theta + "cos" theta)`
`therefore (1)/(r_(2)) + (1)/(r_(3)) = (5 "cos" theta - "sin " theta)/(3(1-k)) + ((5 "sin" theta - "cos" theta))/(9(1-k))`
`= (15 "cos" theta - 3"sin " theta + 5"sin" theta -"cos" theta)/(9(1-k))`
`= (14 "cos" theta + 2"sin " theta)/(9(1-k)) = (2)/(r_(1))`
Clearly, AC, AB and AD are in H.P.