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in triangle pab pa=pb and the area of the triangle is 10 sq u

Answer» PA =\xa0{tex}\\sqrt { ( 1 - x ) ^ { 2 } + ( 2 - y ) ^ { 2 } }{/tex}PB =\xa0{tex}\\sqrt { ( 3 - x ) ^ { 2 } + ( 8 - y ) ^ { 2 } }{/tex}{tex}\\therefore{/tex}\xa0PA = PBLet coordinates of P are (x, y){tex}\\Rightarrow{/tex}\xa0{tex} \\sqrt { ( 1 - x ) ^ { 2 } + ( 2 - y ) ^ { 2 } }{/tex}{tex}= \\sqrt { ( 3 - x ) ^ { 2 } + ( 8 - y ) ^ { 2 } }{/tex}Squaring on both sides,\xa0{tex}\\Rightarrow{/tex}\xa0{tex}(1 - x)^2 + (2 - y)^2 = (3 - x)^2 + (8 - y)^2{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}(1 - x)^2 - (3 - x)^2 = (8 - y)^2 - (2 - y)^2{/tex}{tex}\\Rightarrow{/tex}\xa0(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}\xa0{tex}- 2 \\times 2 ( 2 - x ) = 6 \\times 2 ( 5 - y ){/tex}- 2 + x = 15 - 3yx = 17 - 3y ...(i)Area of\xa0{tex}\\triangle{/tex}PAB = 10 sq units{tex}\\Rightarrow{/tex}\xa0{tex} \\frac { 1 } { 2 } | x ( 2 - 8 ) + 1 ( 8 - y ) + 3 ( y - 2 ) |{/tex}\xa0= 10{tex}\\Rightarrow{/tex}\xa0{tex}| - 6 x + 8 - y + 3 y - 6 |{/tex}\xa0= 20{tex}\\Rightarrow{/tex}\xa0{tex}| - 6 x + 2 y + 2 |{/tex}\xa0= 20{tex}\\Rightarrow{/tex}\xa0{tex}\xa0- 6x + 2y + 2 = 20 {/tex}or\xa0{tex}- 6x + 2y + 2 = -20{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}- 6(17 - 3y) + 2y + 2 = 20{/tex} or {tex}- 6(17 - 3y) + 2y + 2 = - 20 {/tex}...(ii){tex}\\Rightarrow{/tex}\xa0{tex}- 102 + 18y + 2y + 2 = 20{/tex} or\xa0{tex}- 102 + 18y + 2y + 2 = - 20{/tex}{tex}\\Rightarrow{/tex}\xa020y = 120 or 20y = 80y = 6 or y = 4When y = 6, equation (i) becomes x = 17 - 18 = - 1{tex}\\therefore{/tex}\xa0Point is (- 1, 6).When y = 4, equation (i) becomes x = 17 - 12 = 5{tex}\\therefore{/tex}\xa0Point is (5, 4).


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