in triangle pab pa=pb and the …

PA = {tex}\sqrt { ( 1 - x ) ^ { 2 } + ( 2 - y ) ^ { 2 } }{/tex}
PB = {tex}\sqrt { ( 3 - x ) ^ { 2 } + ( 8 - y ) ^ { 2 } }{/tex}
{tex}\therefore{/tex} PA = PB
Let coordinates of P are (x, y)
{tex}\Rightarrow{/tex} {tex} \sqrt { ( 1 - x ) ^ { 2 } + ( 2 - y ) ^ { 2 } }{/tex}{tex}= \sqrt { ( 3 - x ) ^ { 2 } + ( 8 - y ) ^ { 2 } }{/tex}
Squaring on both sides, 
{tex}\Rightarrow{/tex} {tex}(1 - x)^2 + (2 - y)^2 = (3 - x)^2 + (8 - y)^2{/tex}
{tex}\Rightarrow{/tex} {tex}(1 - x)^2 - (3 - x)^2 = (8 - y)^2 - (2 - y)^2{/tex}
{tex}\Rightarrow{/tex} (1 - x - 3 + x) (1 - x + 3 - x) = (8 - y - 2 + y) (8 - y + 2 - y)
{tex}\Rightarrow{/tex}- 2(4 - 2x) = 6(10 - 2y)
{tex}\Rightarrow{/tex} {tex}- 2 \times 2 ( 2 - x ) = 6 \times 2 ( 5 - y ){/tex}
- 2 + x = 15 - 3y
x = 17 - 3y ...(i)
Area of {tex}\triangle{/tex}PAB = 10 sq units
{tex}\Rightarrow{/tex} {tex} \frac { 1 } { 2 } | x ( 2 - 8 ) + 1 ( 8 - y ) + 3 ( y - 2 ) |{/tex} = 10
{tex}\Rightarrow{/tex} {tex}| - 6 x + 8 - y + 3 y - 6 |{/tex} = 20
{tex}\Rightarrow{/tex} {tex}| - 6 x + 2 y + 2 |{/tex} = 20
{tex}\Rightarrow{/tex} {tex} - 6x + 2y + 2 = 20 {/tex}or {tex}- 6x + 2y + 2 = -20{/tex}
{tex}\Rightarrow{/tex} {tex}- 6(17 - 3y) + 2y + 2 = 20{/tex} or {tex}- 6(17 - 3y) + 2y + 2 = - 20 {/tex}...(ii)
{tex}\Rightarrow{/tex} {tex}- 102 + 18y  + 2y + 2 = 20{/tex} or {tex}- 102 + 18y + 2y + 2 = - 20{/tex}
{tex}\Rightarrow{/tex} 20y = 120 or 20y = 80
y = 6 or y = 4
When y = 6, equation (i) becomes x = 17 - 18 = - 1
{tex}\therefore{/tex} Point is (- 1, 6).
When y = 4, equation (i) becomes x = 17 - 12 = 5
{tex}\therefore{/tex} Point is (5, 4).