Exercises on three-point alignment condition

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Lined dots or collinear points they are points that belong to the same line.

Given three points $\dpi{120} \mathrm{A}(x_1,y_1)$ , $\dpi{120} \mathrm{B}(x_2,y_2)$ and $\dpi{120} \mathrm{C}(x_3,y_3)$ , the condition of alignment between them is that the coordinates are proportional:

$\dpi{120} \boldsymbol{\frac{x_2-x_1}{x_3-x_2} = \frac{y_2-y_1}{y_3-y_2}}$

See a list of exercises on three-point alignment condition, all with full resolution.

Index

Exercises on three-point alignment condition
Resolution of question 1
Resolution of question 2
Resolution of question 3
Resolution of question 4
Resolution of question 5

Exercises on three-point alignment condition

Question 1. Check that the points (-4, -3), (-1, 1) and (2, 5) are aligned.

Question 2. Check that the points (-4, 5), (-3, 2) and (-2, -2) are aligned.

Question 3. Check if the points (-5, 3), (-3, 1) and (1, -4) belong to the same line.

Question 4. Determine the value of a so that the points (6, 4), (3, 2) and (a, -2) are collinear.

Question 5. Determine the value of b for the points (1, 4), (3, 1) and (5, b) that are vertices of any triangle.

Resolution of question 1

Points: (-4, -3), (-1, 1) and (2, 5).

We calculate the first side of the equality:

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$\dpi{120} \frac{x_2-x_1}{x_3-x_2} = \frac{-1 - (-4)}{2 - (-1)} = \frac{3}{3} = 1$

We calculate the second side of the equality:

$\dpi{120} \frac{y_2-y_1}{y_3-y_2} = \frac{1 - (-3)}{5 - 1} = \frac{4}{4}=1$

Since the results are equal (1 = 1), then the three points are aligned.

Resolution of question 2

Points: (-4, 5), (-3, 2) and (-2, -2).

We calculate the first side of the equality:

$\dpi{120} \frac{x_2-x_1}{x_3-x_2} = \frac{-3 - (-4)}{-2-(-3)} = \frac{1}{1} = 1$

We calculate the second side of the equality:

$\dpi{120} \frac{y_2-y_1}{y_3-y_2} = \frac{2 - 5}{-2-2} = \frac{-3}{-4}= \frac{3}{4 }$

How the results are different $\bigg (1\neq \frac{3}{4}\bigg)$ , so the three points are not aligned.

Resolution of question 3

Points: (-5, 3), (-3, 1) and (1, -4).

We calculate the first side of the equality:

$\dpi{120} \frac{x_2-x_1}{x_3-x_2} = \frac{-3 - (-5)}{1 - (-3)} = \frac{2}{4} = \frac{ 1}{2}$

We calculate the second side of the equality:

$\dpi{120} \frac{y_2-y_1}{y_3-y_2} = \frac{1 - 3}{-4 - 1} = \frac{-2}{-5}= \frac{2}{5 }$

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How the results are different $\bigg(\frac{1}{2}\neq \frac{2}{5}\bigg)$ , so the three points are not aligned, so they do not belong to the same line.

Resolution of question 4

Points: (6, 4), (3, 2) and (a, -2)

Collinear points are aligned points. So we must get the value of a so that:

$\dpi{120} \frac{x_2-x_1}{x_3-x_2} = \frac{y_2-y_1}{y_3-y_2}$

Substituting the coordinate values, we have to:

$\dpi{120} \mathrm{\frac{3-6}{a-3} = \frac{2-4}{-2-2}}$

$\dpi{120} \Rightarrow \mathrm{\frac{-3}{a-3} = \frac{-2}{-4}}$

Applying the fundamental property of proportions (cross multiplication):

$\dpi{120} \mathrm{-2(a-3)=12}$

$\dpi{120} \Rightarrow \mathrm{-2a + 6=12}$

$\dpi{120} \Rightarrow \mathrm{-2a = 6}$

$\dpi{120} \Rightarrow \mathrm{a = -\frac{6}{2}}$

$\dpi{120} \Rightarrow \mathrm{a = -3}$

Resolution of question 5

Points: (1, 4), (3, 1) and (5, b).

The vertices of a triangle are unaligned points. So let's get the value of b to which the points are aligned and any other different value will result in non-aligned points.

$\dpi{120} \frac{x_2-x_1}{x_3-x_2}= \frac{y_2-y_1}{y_3-y_2}$