Question

If the points (a, 0), (0, b) and (1, 1) are collinear, then which of the following is true?

• A. $\frac{1}{a}$ + $\frac{1}{b}=2$
• B. $\frac{1}{a}$ + $\frac{1}{b}=1$
• C. $\frac{1}{a}$ + $\frac{1}{b}=0$
• D. $\frac{1}{a}$ + $\frac{1}{b}=4$

Answer 1

The correct option is B

$\frac{1}{a}$ + $\frac{1}{b}=1$

For 3 points to be collinear the area of the triangle formed by these three points should be zero.

Area of a triangle formed by $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by
$\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

$\therefore \frac{1}{2}\left[a\right(b-1)+0(1-0)+1(0-b\left)\right]=0\frac{1}{2}\left[a\right(b-1)+1(0-b\left)\right]=0ab=a+b\frac{1}{a}+\frac{1}{b}=1$