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What will be the triangle with vertices $(4, 3), (- 3, 2), (1, - 6)$ called?

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An obtuse-angled triangle

An acute-angled triangle

A right-angled triangle

A right-angled isosceles

answer is B.

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Detailed Solution

Let us consider a triangle

ABC

with angles

∠ A, ∠ B

and

∠ C

.
Now, the points given are

= A (4, 3), B (- 3, 2), C (1, - 6)

.
Let’s find out the slope of two points by using the formula,

\Rightarrow slope = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

And then at last we will be finding the angles by using the formula,

\Rightarrow tan θ = \frac{m_{2} - m_{1}}{1 + m_{2} m_{3}}

Now,
For line AB,
As given,

= A (4, 3), B (- 3, 2)

So, according to the formula given above,

\Rightarrow slope = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

\Rightarrow m_{AB} = \frac{2 - 3}{- 3 - 4}

\Rightarrow m_{AB} = \frac{- 1}{- 7}

\Rightarrow m_{AB} = \frac{1}{7}

For line BC,
As given,

= B (- 3, 2), C (1, - 6)

So, according to the formula given above,

\Rightarrow slope = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

\Rightarrow m_{BC} = \frac{- 6 - 2}{1 - (- 3)}

\Rightarrow m_{BC} = \frac{- 8}{1 + 3}

\Rightarrow m_{BC} = - \frac{8}{4}

\Rightarrow m_{BC} = - 2

For line AC,
As given,

= A (4, 3), C (1, - 6)

So, according to the formula given above,

\Rightarrow slope = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

\Rightarrow m_{AC} = \frac{- 6 - 3}{1 - 4}

\Rightarrow m_{AC} = \frac{- 9}{- 3}

\Rightarrow m_{AC} = 3

Now, we will be using this to find the required angles.
Now, let us assume that between the lines AB and BC the angle is

∠ B

.
Similarly, the angle between BC and AC is

∠ C

.
And, the angle between AB and AC is

∠ A

.
In line AC and BC, According to the formula given above,

\Rightarrow \tan ∠ C = \frac{m_{BC} - m_{AC}}{1 + m_{AC} m_{BC}}

\Rightarrow \tan ∠ C = \frac{- 2 - 3}{1 + (3) (- 2)}

\Rightarrow \tan ∠ C = \frac{- 5}{1 - 6}

\Rightarrow \tan ∠ C = \frac{- 5}{- 5}

\Rightarrow \tan ∠ C = 1

\Rightarrow ∠ C = \tan^{- 1} (1)

\Rightarrow ∠ C = 45^{\circ}

In line AB and BC, According to the formula given above,

\Rightarrow \tan ∠ B = \frac{m_{AB} - m_{BC}}{1 + m_{AB} m_{BC}}

\Rightarrow \tan ∠ B = \frac{\frac{1}{7} - (- 2)}{1 + (\frac{1}{7}) (- 2)}

\Rightarrow \tan ∠ B = \frac{\frac{1 + 14}{7}}{1 - \frac{2}{7}}

\Rightarrow \tan ∠ B = \frac{\frac{15}{7}}{\frac{7 - 2}{7}}

\Rightarrow \tan ∠ B = \frac{\frac{15}{7}}{\frac{5}{7}}

\Rightarrow \tan ∠ B = \frac{15}{5}

\Rightarrow \tan ∠ B = 3

\Rightarrow ∠ B = \tan^{- 1} (3)

\Rightarrow ∠ B = {71.56}^{\circ}

In line AB and AC, According to the formula given above,

\Rightarrow \tan ∠ A = \frac{m_{AC} - m_{AB}}{1 + m_{AC} m_{AB}}

\Rightarrow \tan ∠ A = \frac{3 - \frac{1}{7}}{1 + (3) (\frac{1}{7})}

\Rightarrow \tan ∠ A = \frac{\frac{21 - 1}{7}}{1 + \frac{3}{7}}

\Rightarrow \tan ∠ A = \frac{\frac{20}{7}}{\frac{7 + 3}{7}}

\Rightarrow \tan ∠ A = \frac{20}{10}

\Rightarrow ∠ A = \tan^{- 1} (2)

\Rightarrow ∠ A = {63.43}^{\circ}

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What will be the triangle with vertices $(4, 3), (- 3, 2), (1, - 6)$ called?