 QUESTION

# If the points (a, 1), (1, b) and (a-1, b-1) are collinear. $\alpha ,\beta$ are respectively the arithmetic and geometric means of a and b, then find the value of $4\alpha - {\beta ^2}$.A.-1B.0C.3D.2

Hint: Make use of the properties of the determinants which says that if three given points are collinear, then the value of the determinant is zero.

The given points are (a, 1), (1, b) and (a-1, b-1). And said these points are collinear.
If points $\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right),\left( {{x_3},{y_3}} \right)$ are collinear, then $\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1 \\ {{x_2}}&{{y_2}}&1 \\ {{x_3}}&{{y_3}}&1 \end{array}} \right|$
So for the given points, $\left| {\begin{array}{*{20}{c}} a&1&1 \\ 1&b&1 \\ {a - 1}&{b - 1}&1 \end{array}} \right| = 0$
On expanding we get
$a(b - b + 1) - 1(1 - a + 1) + 1(b - 1 - ab + b) = 0$
$2(a + b) - ab = 3$ .... (1)
From the given information
$\begin{gathered} \alpha = \dfrac{{a + b}}{2}\;....(2) \\ \;\beta = \sqrt {ab} \;....(3) \\ \end{gathered}$
Substituting the values of $\alpha ,\beta$ from equations (2), (3)in equation (1), to convert that into $\alpha ,\beta$.
$\Rightarrow 2(2\alpha ) - {\beta ^2} = 3$
$4\alpha - {\beta ^2} = 3$
$\therefore$ The value of $4\alpha - {\beta ^2} = 3$.

Note:Here we used the determinant of the matrix for assuming the points are collinear. We can also use slopes of any two pair of points to make it.Three points are collinear if and only if the determinant found by placing the x-coordinates in the first column, the y-coordinates in the second column, and one's in the third column is equal to zero.