Question

Determine if the points (1,5).(2.3) and (-2,-11) are collinear.
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Answer 1

There are not collinear.

Explanation:

Let us name the points as A(1,5), B(2,3) and C(-2,-11) respectively.

For these points to be collinear, we need to have:

AB + BC = AC

(The sum of the lengths of the two small segments must be equal to the bigger line segment.

Now, finding the lengths,

AB --- A(1,5) and B(2,3)

$\sf{\longrightarrow\:AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$

$\sf{\longrightarrow\:AB=\sqrt{(2-1)^2+(3-5)^2}}$

$\sf{\longrightarrow\:AB=\sqrt{1+4}}$

$\sf{\longrightarrow\:AB=\sqrt{5}}$

BC --- B(2,3) and C(-2,-11)

$\sf{\longrightarrow\:BC=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$

$\sf{\longrightarrow\:AB=\sqrt{(-2-2)^2+(-11-3)^2}}$

$\sf{\longrightarrow\:AB=\sqrt{(-4)^2+(-14)^2}}$

$\sf{\longrightarrow\:AB=\sqrt{16+196}}$

$\sf{\longrightarrow\:AB=\sqrt{212}}$

AC --- A(1,5) and C(-2,-11)

$\sf{\longrightarrow\:AC=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$

$\sf{\longrightarrow\:AC=\sqrt{(-2-1)^2+(-11-5)^2}}$

$\sf{\longrightarrow\:AC=\sqrt{(-3)^2+(-16)^2}}$

$\sf{\longrightarrow\:AC=\sqrt{9+256}}$

$\sf{\longrightarrow\:AB=\sqrt{265}}$

Here, we can infer that the sum of root of 212 and root of 5 is not equal to root of 265 which means that the given points are not collinear.

(AB + BC ≠ AC)

Answer 2

(Alternative Method)

Answer:

The three points are not collinear.

Explanation:

Three points are given.

If we calculate the area of triangle formed by these thee points, and the area is equal to 0, we can say that these points are collinear.

$\text{Area of triangle = } \left| \dfrac{1}{2} \left[ x_1(y_2 - y_3) + x_2(y_3 - y_1) + x3(y1 - y2) \right] \right|\\\\\\=\dfrac{1}{2} \times \left| [1(3-(-11)+2(-11-5)+(-2)(5-3)] \right|\\\\\\= \dfrac{1}{2} \times \left| [14 - 32 - 4] \right|\\\\\\= \dfrac{1}{2} \times |[-22]|\\\\\\= 22 \div 2\\\\\\= 11 \neq 0$

Clearly, the area of the triangle is not equal to 0.

Therefore, the three points are not collinear.