Question

Do the points A (3, 2), B (-2, -3) and C (2, 3) form a triangle? If so, name the type of triangle formed.

Answer 1

Hint: Here, first we have to apply the distance formula, $d=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}$ to get the values of AB, BC and AC. Then, we will get the square of BC as the sum of the squares of AB and AC. Then with the help of Pythagoras theorem we can find the nature of the triangle.

Complete step-by-step answer:

Hence, we are given with the points A (3, 2), B (-2, -3) and C (2, 3).

Now, we have to check whether these vertices are the vertices of a triangle or not.

For that first we have to use the distance formula. Let $P({{x}_{1}},{{y}_{1}})$ and $Q({{x}_{2}},{{y}_{2}})$ be two points then the distance between P and Q is given by the distance formula:

$PQ=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}$

By applying the distance formula we have to find the values of AB, BC and AC.

First we can find AB. We have A (3, 2) and B (-2 -3) where $({{x}_{1}},{{y}_{1}})=(3,2)$ and $({{x}_{2}},{{y}_{2}})=(-2,-3)$. Hence, we will get:

$ AB=\sqrt{{{(-2-3)}^{2}}+{{(-3-2)}^{2}}} $

$  AB=\sqrt{{{(-5)}^{2}}+{{(-5)}^{2}}} $

$  AB=\sqrt{25+25} $

$  AB=\sqrt{50} $

Now, we can factorise 50, we have:

$50=2\times 5\times 5$

Now, AB can be written as:

$ AB=\sqrt{2\times 5\times 5} $

$  AB=\sqrt{2\times {{5}^{2}}} $

$  AB=\sqrt{2}\times \sqrt{{{5}^{2}}} $

$  AB=\sqrt{2}\times 5 $

$  AB=5\sqrt{2} $

Hence, we got the value of $AB=5\sqrt{2}$.

Now, we have to find the value of BC, B (-2, -3) and C (2, 3) where $({{x}_{1}},{{y}_{1}})=(-2,-3)$ and $({{x}_{2}},{{y}_{2}})=(2,3)$. Hence we obtain:

$  BC=\sqrt{{{(2-(-2))}^{2}}+{{(3-(-3))}^{2}}} $

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

$  BC=\sqrt{{{4}^{2}}+{{6}^{2}}} $

$ BC=\sqrt{16+36} $

$ BC=\sqrt{52} $

Now, we can factorise 52, we will get:

$52=13\times 2\times 2$

Now, $AB$ can be written as:

$  BC=\sqrt{13\times 2\times 2} $

$  BC=\sqrt{13\times {{2}^{2}}} $

$  BC=\sqrt{13}\times \sqrt{{{2}^{2}}} $

$  BC=\sqrt{13}\times 2 $

$ BC=2\sqrt{13} $

Next, we have to find the value of AC. We have A (3, 2) and C (2, 3) where $({{x}_{1}},{{y}_{1}})=(3,2)$ and $({{x}_{2}},{{y}_{2}})=(2,3)$.Hence we obtain:

$ AC=\sqrt{{{(2-3)}^{2}}+{{(3-2)}^{2}}} $

$  AC=\sqrt{{{-1}^{2}}+{{(1)}^{2}}} $

$ AC=\sqrt{1+1} $

$  AC=\sqrt{2} $

Now let us take the square of AB, BC and AC, we will get:

$  AB=\sqrt{50} $

$  A{{B}^{2}}={{\left( \sqrt{50} \right)}^{2}} $

$  A{{B}^{2}}=50 $

Similarly, we will obtain:

$  BC=\sqrt{52} $

$  B{{C}^{2}}={{\left( \sqrt{52} \right)}^{2}} $

$ B{{C}^{2}}=52 $

Next, we will get:

$ AC=\sqrt{2}$

$ A{{C}^{2}}={{\left( \sqrt{2} \right)}^{2}} $

$ A{{C}^{2}}=2 $

Now, from the above data we can write:

$52=50+2$

Hence, we can write:

$B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}$

The above equation satisfies the Pythagoras theorem which says that in a right angled triangle the square of the hypotenuse is the sum of the squares of its base and altitude. Here, BC is the hypotenuse, AB is the altitude and AC is the base of the triangle $\Delta ABC$.

We know that Pythagoras theorem is only applicable for a right angled triangle. Therefore, by Pythagoras theorem, we can say that the triangle $\Delta ABC$ is a right angled triangle.

Hence, we can say that the points A (3, 2), B (-2, -3) and C (2, 3) form the vertices of a right angled triangle.

Note: Here, you can also say that if the sum of any two sides is greater than the third side, then the points are the vertices of the triangle. Clearly this condition is satisfied here.