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| 1. |
Find the circumcentre of the triangle whose vertices are (-2,-3), (-1,0), (7,-6). |
| Answer» Let the coordinates of the circumcentre of the triangle be P(x, y). Then, we know that circumentre of a triangle is equidistant from each of its vertices.\xa0{tex}PA = \\sqrt {{{(x + 2)}^2} + {{(y + 3)}^2}} {/tex}{tex} \\Rightarrow P{A^2} = {x^2} + {y^2} + 4x + 6y + 13{/tex}\xa0..... (i){tex} \\Rightarrow PB = \\sqrt {{{(x + 1)}^2} + {{(y - 0)}^2}} {/tex}{tex} \\Rightarrow P{B^2} = {x^2} + {y^2} + 2x + 1{/tex}\xa0...... (ii){tex}PC = \\sqrt {{{(x - 7)}^2} + {{(y + 6)}^2}} {/tex}{tex} \\Rightarrow {/tex}\xa0PC2 = x2 + y2 - 14x + 12y + 85 ..... (iii)Now, PA2 = PB2{tex} \\Rightarrow {/tex}\xa0x2 + y2 + 4x + 6y + 13 = x2 + y2 + 2x + 1{tex} \\Rightarrow {/tex}\xa02x + 6y = -12{tex} \\Rightarrow {/tex}\xa0x + 3y = -6 ..... (iv)And PA2 = PC2x2 + y2 + 4x + 6y + 13 = x2 + y2 - 14x + 12y + 8518x - 6y = 72{tex} \\Rightarrow {/tex} 3x - y = 12 ..... (v)And PB2 = PC2{tex} \\Rightarrow {/tex}\xa0x2 + y2 + 2x + 1 = x2 + y2 - 14x + 12y + 85{tex} \\Rightarrow {/tex}\xa016x - 12y = 84\xa0{tex} \\Rightarrow {/tex}\xa04x - 3y = 21 ....... (vi)Solving (iv) and (v) we get x = 3, y = -3Hence, circumcentre of the triangle is (3, -3) | |