Find the vertices of triangle, the mid points of whose sides are (3,1)

1.	Find the vertices of triangle, the mid points of whose sides are (3,1), (5,6) and (-3,2).
Answer» Let vertices of {tex} \\triangle ABC{/tex}\xa0be A(x1,\xa0y1), B(x2, y2) and C(x3, y3)By mid-points formula{tex}\\frac{{{x_2} + {x_3}}}{2} = 3 \\Rightarrow {x_2} + {x_3} = 6{/tex}\xa0...... (i){tex}\\frac{{{y_2} + {y_3}}}{2} = 1 \\Rightarrow {y_2} + {y_3} = 2{/tex}\xa0...... (ii){tex}\\frac{{{x_3} + {x_1}}}{2} = 5 \\Rightarrow {x_3} + {x_1} = 10{/tex}\xa0..... (iii){tex}\\frac{{{y_3} + {y_1}}}{2} = 6 \\Rightarrow {y_1} + {y_3} = 12{/tex}\xa0..... (iv){tex}\\frac{{{x_1} + {x_2}}}{2} = - 3 \\Rightarrow {x_1} + {x_2} = - 6{/tex}\xa0..... (v){tex}\\frac{{{y_1} + {y_2}}}{2} = 2 \\Rightarrow {y_1} + {y_2} = 4{/tex}\xa0...... (vi)Adding (i), (iii) and (v)2(x1 + x2 + x3) = 10{tex} \\Rightarrow {/tex}\xa0x1 + x2 + x3 = 5 ...... (vii)Adding (ii),(iv)and (vi)2(y1 + y2 + y3) = 18y1 + y2 + y3 = 9 ....... (viii)Subtracting (i), (iii) and (v) from (vii)We get, x1 = -1, x2 = -5, x3 = 11Subtracting (ii), (iv) and (vi) from eq. (viii)We get, y1 = 7, y2 = -3, y3 = 5

Find the vertices of triangle, the mid points of whose sides are (3,1), (5,6) and (-3,2).