Question

Prove that the points (a, -3a), (2a, a) and (0, -2a) form a scalene trqingle​

Answer 1

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Answer 2

Answer:

See below.

Step-by-step explanation:

Remember that a scalene triangle has lengths of different values.

Therefore, we just need to find the length or distance from each point to the next. If the three distances we acquire are different, then we prove that the point do indeed form a scalene triangle.

Let's let A be (a, -3a), B be (2a, a), and C be (0, -2a).

So, let's find each of the side lengths using the distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2$

Side AB:

Let's let A:(a, -3a) be (x₁, y₁) and let's let B:(2a, a) be (x₂, y₂). Substitute this into our formula:

$d=\sqrt{(2a-a)^2+(a-(-3a))^2$

Subtract:

$d=\sqrt{(a)^2+(4a)^2$

Square:

$d=\sqrt{a^2+16a^2}$

Add:

$d=\sqrt{17a^2}$

Simplify:

$d=\sqrt{a^2}\cdot \sqrt{17}\\d=|a|\sqrt{17}$

So:

$\overline {AB}=|a|\sqrt{17}$

Note: We need the absolute value because anything squared will be positive, and if you take the square root of something positive, the result will be positive. The absolute value ensures that the a value will be positive no matter what a is to begin with.

Side BC:

Let's let C:(0, -2a) be (x₁, y₁) and let's let B:(2a, a) be (x₂, y₂).

$d=\sqrt{(2a-0)^2+(a-(-2a)^2}$

Subtract:

$d=\sqrt{(2a)^2+(3a)^2}$

Square:

$d=\sqrt{4a^2+9a^2}$

Add:

$d=\sqrt{13a^2}$

Simplify:

$d=\sqrt{a^2}\cdot \sqrt{13}\\d=|a|\sqrt{13}$

Therefore:

$\overline{BC}=|a|\sqrt{13}$

Side AC:

Let's let A:(a, -3a) be (x₁, y₁) and let's let C:(0, -2a) be (x₂, y₂).

$d=\sqrt{(0-a)^2+(-2a-(-3a))^2$

Subtract:

$d=\sqrt{(-a)^2+(a)^2}$

Square:

$d=\sqrt{a^2+a^2}$

Add:

$d=\sqrt{2a^2}$

Simplify:

$d=|a|\sqrt2$

Therefore:

$\overline{AC}=|a|\sqrt2$

So, our three side lengths are:

$\overline {AB}=|a|\sqrt{17}\text{, }\overline{BC}=|a|\sqrt{13}\text{, and } \overline{AC}=|a|\sqrt2$

We can see that the three side lengths are different since they do not equal to same thing.

Therefore, we can deduce that the triangle must be scalene.

And we're done!