Math / Geometry

Question9. দেখাও যে, $A(0,-3), B(4,-2)$ এবং $C(16,1)$ বिन্দু তিনটি সমরেখ।

Studdy Solution

STEP 1

1. The coordinates of the points $A$ , $B$ , and $C$ are given as $A(0,-3)$ , $B(4,-2)$ , and $C(16,1)$ respectively.
2. Points $A$ , $B$ , and $C$ are collinear if the area of the triangle formed by these three points is zero.
3. The area of a triangle with vertices $(x_1, y_1)$ , $(x_2, y_2)$ , $(x_3, y_3)$ can be calculated using the determinant formula.

STEP 2

1. Substitute the coordinates of the points $A$ , $B$ , and $C$ into the determinant formula for the area of a triangle.
2. Simplify the determinant to determine whether the area is zero.
3. Conclude whether the points are collinear based on the area.

STEP 3

Write the determinant formula for the area of the triangle formed by points $A(0, -3)$ , $B(4, -2)$ , and $C(16, 1)$ .
The area $\Delta$ is given by:
$\Delta = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$ where $(x_1, y_1) = (0, -3)$ , $(x_2, y_2) = (4, -2)$ , $(x_3, y_3) = (16, 1)$ .

STEP 4

Substitute the coordinates of points $A$ , $B$ , and $C$ into the formula.
$\Delta = \frac{1}{2} \left| 0(-2 - 1) + 4(1 - (-3)) + 16((-3) - (-2)) \right|$

STEP 5

Simplify the expression inside the absolute value.
$\Delta = \frac{1}{2} \left| 0(-3) + 4(1 + 3) + 16(-3 + 2) \right|$ $\Delta = \frac{1}{2} \left| 0 + 4 \cdot 4 + 16(-1) \right|$ $\Delta = \frac{1}{2} \left| 0 + 16 - 16 \right|$ $\Delta = \frac{1}{2} \left| 0 \right|$ $\Delta = \frac{1}{2} \cdot 0$ $\Delta = 0$

STEP 6

Since the area $\Delta$ is $0$ , the points $A$ , $B$ , and $C$ are collinear.

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