Question

given prove that the points (a, a square),(b,b square,).. c) can never be collinear.​

Answer 1

Let A(a,a²), B(b,b²), C(0,0) be the coordinates of the given points.

We know that the area of a triangle having vertices (x1,y1), (x2,y2),(x3,y3) is [½(x1(y2-y3)+x2(y3-y1)+x3(y1-y2) Square units.

Area of ∆ABC= [½(x1(y2-y3)+x2(y3-y1)+x3(y1-y2) Square units.

= |½[(a(b²-0)+b(0-a²)+0(a²-b²)]|

= |½(ab²-0+0+a²b+0-0)|

= |½(ab²+a²b)|

≠0  (a≠b≠0) given

Since the area of the triangle formed by the points (a, a²),(b,b²),(0,0) is not zero ,so the given points are not collinear

Answer 2

Answer: cyka

Explanation:

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