if a not= b not= 0 prove that (a,a×a) (b, b×b) (0, 0) will not collinear
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SOLUTION: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.
i hope this is what your asking for ??
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.
i hope this is what your asking for ??
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