Question

PLS HELP BRAINLEIST AND 25 POINTS

Answer 1

Answer:

Step-by-step explanation:

A( 0, 2), B (2, 5), and C (−1, 7)

(A) The distance formula is given by

$d = \sqrt {(x'-x)^2+(y'-y)^2}$

length of side AB

$AB= \sqrt {(2 - 0)^2+(5 - 2)^2}=\sqrt {13}$

Length of side BC

$BC= \sqrt {(- 1 - 2)^2+(7 - 5)^2}=\sqrt {13}$

Length of side CA

$CA= \sqrt {(- 1 - 0)^2+(7 - 2)^2}=\sqrt {26}$

(B) The slope of a line is given by

$m = \frac{y' - y}{x' - x}$

Slope of line AB = $m = \frac{5-2}{2-0}=\frac{3}{2}$

Slope of line BC = $m = \frac{7 - 5}{-1 - 2}=-\frac{2}{3}$

Slope of line CA = $m = \frac{7 - 2}{-1 - 0}=- 5$

(c) As the slope of line AB is negative of reciprocal of line BC, and length AB is equal to length BC so the triangle is right angled isosceles triangle.

Answer 2

Given:

The vertices of a triangle ABC are A( 0, 2), B (2, 5), and C (−1, 7).

To find:

Part A: The length of each side of the triangle.

Part B: The slope of each side of the triangle.

Part C: Classify the triangle.

Solution:

Part A: The vertices of a triangle ABC are A( 0, 2), B (2, 5), and C (−1, 7).

Distance formula:

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

Using the distance formula, we get

$AB=\sqrt{(2-0)^2+(5-2)^2}$

$AB=\sqrt{2^2+3^2}$

$AB=\sqrt{4+9}$

$AB=\sqrt{13}$

Similarly,

$BC=\sqrt{\left(-1-2\right)^2+\left(7-5\right)^2}$

$BC=\sqrt{13}$

And,

$AC=\sqrt{\left(-1-0\right)^2+\left(7-2\right)^2}$

$AC=\sqrt{26}$

Therefore, the length of each side of the triangle are $AB=\sqrt{13},BC=\sqrt{13},AC=\sqrt{26}$.

Part B:

Slope formula:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

Slope of side AB is:

$m_{AB}=\dfrac{5-2}{2-0}$

$m_{AB}=\dfrac{3}{2}$

Slope of side BC is:

$m_{BC}=\dfrac{7-5}{-1-2}$

$m_{BC}=\dfrac{2}{-3}$

$m_{BC}=-\dfrac{2}{3}$

Slope of side AC is:

$m_{AC}=\dfrac{7-2}{-1-0}$

$m_{AC}=\dfrac{5}{-1}$

$m_{AC}=-5$

Therefore, the slopes of sides AB, BC, AC are $\dfrac{3}{2},-\dfrac{2}{3},-5$ respectively.

Part C:

$AB=BC=\sqrt{13}$

Product of slopes of AB and BC is:

$\dfrac{3}{2}\times \dfrac{-2}{3}=-1$

So, side AB and BC are perpendicular to each other.

Two sides of triangle ABC are equal and one angle is a right angle.

Therefore, the triangle ABC is an isosceles right triangle.