Question

For what value of a the points a,11 ; 1,-1 and 11,4 are collinear?

Answer 1

Corrected question: For what value of a the points (a, 1), (1, -1) and (11 ,4) are collinear?

Answer:

The value of 'a' is 5.

Step-by-step explanation:

ATQ, the points (a, 1), (1, -1) & (11 4) are collinear.

We know that when three collinear points are joined using straight lines, the area enclosed by them will be equal to 0.

Therefore, we'll use the Area of a triangle formed by coordinates formula to find out the value of 'a'.

$\boxed{\boxed{\sf \ Area \ of \ \triangle = \frac{1}{2} \Bigg( \ x_1\left(y_2 - y_3\right) + x_2\left(y_3 - y_1\right) + x_3\left(y_1 - y_2\right)\Bigg) \ }}$

ATQ,

x₁ $\longmapsto$ a

x₂ $\longmapsto$ 1

x₃ $\longmapsto$ 11

y₁ $\longmapsto$ 11

y₂ $\longmapsto$ -1

y₃ $\longmapsto$ 4

Since we already know the area enclosed by these three points is 0, we'll directly equate it with the area of a triangle formula.

$\Longrightarrow \sf Area \ of \ \triangle = 0\\ \\ \\\Longrightarrow \sf \dfrac{1}{2} \Bigg(x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\Bigg)} = 0\\ \\ \\\Longrightarrow \sf a(-1 - 4)) + 1(4 - 1) + 11(1 - (-1)) = 0 \times 2\\ \\ \\\Longrightarrow \sf (a(-5) + 1(3) + 11(1 + 1) = 0$

$\Longrightarrow \sf -5a + 3 + 11(2) = 0\\ \\ \\\Longrightarrow \sf -5a + 3 + 22 = 0\\ \\ \\\Longrightarrow \sf -5a + 25 = 0\\ \\ \\\Longrightarrow \sf 5a = 25\\ \\ \\\Longrightarrow \sf a = 5$

∴ The value of 'a' is 5.