Question

24. If a *b #0, prove that (a, a?), (b, b?), (0,0) will not be collinear.​

Answer 1

$\large\bold{\underline{\underline{Answer:-}}}$

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

We know that the area of triangle having vertices (x¹, y¹), (x², y²) and (x³, y³) is ∣∣ 12

[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]∣∣ square units.

So,

Area of ∆ABC

=|12|a(b2-0)+b(0-a2)+0(a2- b2)∣∣

=|12(ab2-a2b)|

=12|ab(b-a)|

≠0 (∵a≠b≠0)

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

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Answer 2

Answer:

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

We know that the area of triangle having vertices (x¹, y¹), (x², y²) and (x³, y³) is ∣∣ 12

[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]∣∣ square units.

So,

Area of ∆ABC

=|12|a(b2-0)+b(0-a2)+0(a2- b2)∣∣

=|12(ab2-a2b)|

=12|ab(b-a)|

≠0 (∵a≠b≠0)

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

Step-by-step explanation:

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