The $$mid-points$$ of the sides of a triangle are (5$$,$$7$$ ,$$11), (0$$, $$8$$, $$5) and (2$$,$$3$$, $$-$$ $$1). Then, the vertices are

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Infinity Learn · Accepted Answer

Let the vertices of a triangle are $$Ax1$$,$$y1$$,$$z1$$,$$Bx2$$,$$y2$$,$$z2$$ and $$Cx3$$,$$y3$$,$$z3$$. Since, D, E and F are the $$mid-points$$ of AC ,BC and AB∴$$x1+x22$$,$$y1+y22$$,$$z1+z22=(0$$,$$8$$,$$5)$$⇒$$x1+x2=0$$,$$y1+y2=16$$,$$z1+z2=10-----i$$ $$x2+x32$$,$$y2+y32$$,$$z2+z32=(2$$,$$3$$,−$$1)$$⇒$$x2+x3=4$$,$$y2+y3=6$$,$$z2+z3=$$−$$2----ii$$ and $$x1+x32$$,$$y1+y32$$,$$z1+z32=(5$$,$$7$$,$$11)$$⇒$$x1+x3=10$$,$$y1+y3=14$$,$$z1+z3=22----iiiOn$$ adding Eqs. (i), (ii) and (iir), we $$get2x1+x2+x3=14$$,$$2y1+y2+y3=36$$,$$2z1+z2+z3=30$$⇒$$x1+x2+x3=7$$,$$y1+y2+y3=18$$,$$z1+z2+z3=15$$… (iv) On solving Eqs. (i), (ii), (iil) and (iv)' we $$getx3=7$$,$$x1=3$$,$$x2=$$−$$3y3=2$$,$$y1=12$$,$$y2=4and$$  $$z3=5$$,$$z1=17$$,$$z2=$$−$$7Hence$$, vertices of a triangle are (7$$, $$2$$,$$5), (3$$, $$2$$, $$17), $$(-3$$, $$4$$, $$-7)$$

The mid-points of the sides of a triangle are (5,7 ,11), (0, 8, 5) and (2,3, - 1). Then, the vertices are

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