The points do not form a triangle when they are collinear. To check for collinearity, we can use the determinant formula. For three points \( A(x_1, y_1), B(x_2, y_2), C(x_3, y_3) \), they are collinear if the following determinant is zero: \[ \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0 \] Substitute the coordinates: \[ \begin{vmatrix} a & 2 & 1 \\ 3 & 1 & 1 \\ -1 & 3 & 1 \end{vmatrix} = 0 \] Expanding the determinant: \[ a(1 \cdot 1 - 1 \cdot 3) - 2(3 \cdot 1 - (-1) \cdot 1) + 1(3 \cdot 3 - (-1) \cdot 1) \] \[ = a(1 - 3) - 2(3 + 1) + (9 + 1) \] \[ = a(-2) - 2(4) + 10 = 0 \] \[ -2a - 8 + 10 = 0 \Rightarrow -2a + 2 = 0 \Rightarrow a = 1 \]
In the adjoining figure, \(PQ \parallel XY \parallel BC\), \(AP=2\ \text{cm}, PX=1.5\ \text{cm}, BX=4\ \text{cm}\). If \(QY=0.75\ \text{cm}\), then \(AQ+CY =\)
In the adjoining figure, \( \triangle CAB \) is a right triangle, right angled at A and \( AD \perp BC \). Prove that \( \triangle ADB \sim \triangle CDA \). Further, if \( BC = 10 \text{ cm} \) and \( CD = 2 \text{ cm} \), find the length of } \( AD \).
If a line drawn parallel to one side of a triangle intersecting the other two sides in distinct points divides the two sides in the same ratio, then it is parallel to the third side. State and prove the converse of the above statement.