lf there are three points $$A(2$$,$$3$$, $$4)$$, $$B(-$$ $$1$$,$$2$$, $$-$$ $$3)$$ and $$C(-4$$,$$1$$, $$-$$ $$10)$$ in a space, then they are

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Infinity Learn · Accepted Answer

Given, points are $$A(2$$,$$3$$,$$4)$$, $$B(-1$$,$$2$$, $$-3)$$ and $$C(-4$$,$$1$$, $$-10)$$.Now, determine area of triangleArea of triangle  Δ$$=$$Δ$$xy2+$$Δγ$$z2+$$Δ$$zx2Where$$, Δ$$xy=12x1$$    $$y1$$    $$1x2$$    $$y2$$    $$1x3$$    $$y3$$    $$1$$ , Δ$$yz=12y1$$    $$z1$$    $$1y2$$    $$z2$$    $$1y3$$    $$z3$$    $$1$$  and Δ$$zx=12z1$$    $$x1$$    $$1z2$$    $$x2$$    $$1z3$$    $$x3$$    $$1now$$,Δ$$xy=12231$$−$$121$$−$$411=122(2$$−$$1)$$−$$3($$−$$1+4)$$]$$+1($$−$$1+8)=12$$|[$$2$$−$$9+7$$]|$$=0$$Δ$$yz=123412$$−$$311$$−$$101=123($$−$$3+10)$$−$$4(2$$−$$1)+1($$−$$20+3)=12$$|$$21$$−$$4$$−$$17$$|$$=0$$Δ$$zx=12421$$−$$3$$−$$11$$−$$10$$−$$41=124($$−$$1+4)$$−$$2($$−$$3+10)+1(12$$−$$10)=12$$|[$$12$$−$$14+2$$]|$$=0$$Δ$$=02+02+02=0Hence$$, given points are collinear

lf there are three points A(2,3, 4), B(- 1,2, - 3) and C(-4,1, - 10) in a space, then they are

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