Question

2. The coordinates of the vertices of ABC are A (5, 2), B (2, 4), and C (7, 5). Determine whether ABC is a right triangle.

Answer 1

Answer:

Part 2) Triangle ABC is a right triangle (see the explanation)

Part 3) Quadrilateral QRST is not a parallelogram (see the explanation)

Step-by-step explanation:

Part 2) we have

A (5, 2), B (2, 4), and C (7, 5)

Plot the figure to better understand the problem

see the attached figure

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 1

Find the length side AB

A (5, 2), B (2, 4)

substitute the values in the formula

$d=\sqrt{(4-2)^{2}+(2-5)^{2}}$

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

$d_A_B=\sqrt{13}\ units$

step 2

Find the length side BC

B (2, 4), C (7, 5)

substitute the values in the formula

$d=\sqrt{(5-4)^{2}+(7-2)^{2}}$

$d=\sqrt{(1)^{2}+(5)^{2}}$

$d_B_C=\sqrt{26}\ units$

step 3

Find the length side AC

A (5, 2), C (7, 5)

substitute the values in the formula

$d=\sqrt{(5-2)^{2}+(7-5)^{2}}$

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

$d_A_C=\sqrt{13}\ units$

step 4

Verify if the triangle ABC is a right triangle

we know that a right triangle must satisfy the Pythagorean Theorem

so

$BC^2=AB^2+AC^2$

Remember that the hypotenuse is the greater side

substitute the values

$(\sqrt{26})^2=(\sqrt{13})^2+(\sqrt{13})^2$

$26=13+13$

$26=26$ ----> is true

therefore

Triangle ABC is a right triangle

Part 3) we have

Q (5, 1), R (8, 7), S (14, 10) and T (10, 2)

we know that

The opposite sides of a parallelogram are parallel and congruent

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Step 1

Find the length side QR

Q (5, 1), R (8, 7)

substitute in the formula

$d=\sqrt{(7-1)^{2}+(8-5)^{2}}$

$d=\sqrt{(6)^{2}+(3)^{2}}$

$d_Q_R=\sqrt{45}\ units$

Step 2

Find the length side RS

R (8, 7), S (14, 10)

substitute in the formula

$d=\sqrt{(10-7)^{2}+(14-8)^{2}}$

$d=\sqrt{(3)^{2}+(6)^{2}}$

$d_R_S=\sqrt{45}\ units$

Step 3

Find the length side ST

S (14, 10), T (10, 2)

substitute in the formula

$d=\sqrt{(2-10)^{2}+(10-14)^{2}}$

$d=\sqrt{(-8)^{2}+(-4)^{2}}$

$d_S_T=\sqrt{80}\ units$

Step 4

Find the length side QT

Q (5, 1), T (10, 2)

substitute in the formula

$d=\sqrt{(2-1)^{2}+(10-5)^{2}}$

$d=\sqrt{(1)^{2}+(5)^{2}}$

$d_Q_T=\sqrt{26}\ units$

Step 5

Compare the length of the opposite sides

QR and ST

$\sqrt{45}\ units \neq \sqrt{80}\ units$

$d_Q_R \neq d_S_T$

RS and QT

$\sqrt{45}\ units \neq \sqrt{26}\ units$

$d_R_S \neq d_Q_T$

Opposite sides are not congruent

therefore

Quadrilateral QRST is not a parallelogram