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If the points `(x_1, y_1),(x_2,y_2),` and `(x_3, y_3)` are collinear show that `(y_2-y_3)/(x_2x_3)+(y_3-y_1)/(x_3x_1)+(y_1-y_2)/(x_1x_2)=0`

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STATEMENT-1: If three points (x_(1),y_(1)),(x_(2),y_(2)),(x_(3),y_(3)) are collinear, then |{:(x_(1),y_(1),1),(x_(2),y_(2),1),(x_(3),y_(3),1):}|=0 STATEMENT-2: If |{:(x_(1),y_(1),1),(x_(2),y_(2),1),(x_(3),y_(3),1):}|=0 then the points (x_(1),y_(1)),(x_(2),y_(2)),(x_(3),y_(3)) will be collinear. STATEMENT-3: If lines a_(1)x+b_(1)y+c_(1)=0,a_(2)=0and a_(3)x+b_(3)y+c_(3)=0 are concurrent then |{:(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3)):}|=0

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A triangle has vertices A_i(x_i , y_i)fori=1,2,3 If the orthocentre of triangle is (0,0), then prove that |x_2-x_3 y_2-y_3 y_1(y_2-y_3)+x_1(x_2-x_3) x_3-x_1y_2-y_3y_2(y_3-y_1)+x_1(x_3-x_1) x_1-x_2y_2-y_3y_3(y_1-y_2)+x_1(x_1-x_2)|=0

If the circle x^2+y^2=a^2 intersects the hyperbola x y=c^2 at four points P(x_1, y_1),Q(x_2, y_2),R(x_3, y_3), and S(x_4, y_4), then x_1+x_2+x_3+x_4=0 y_1+y_2+y_3+y_4=0 x_1x_2x_3x_4=C^4 y_1y_2y_3y_4=C^4

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