Question

Match each pair of points A and B to point C such that ∠ABC = 90°. A(3, 3) and B(12, 6) C(6, 52) A(-10, 5) and B(12, 16) C(16, -6) A(-8, 3) and B(12, 8) C(18, 4) A(12, -14) and B(-16, 21) C(-11, 25) A(-12, -19) and B(20, 45) A(30, 20) and B(-20, -15) arrowBoth

Answer 1

Answer:

The pair of points A and B to point C such that ∠ABC = 90° are;

1) A(-10, 5) and B(12, 16), C(18, 4)

2) A(12, -14) and B(-16, 21), C(-11, 25)

Step-by-step explanation:

The angle formed between the points A and B to point C, ∠ABC is 90° when the the line passing through points A and B is perpendicular to the line passing through points B and C

Therefore, for ∠ABC = 90°

$\overline{AB}$ is perpendicular to $\overline{BC}$ (

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

The slope of  $\overline{BC}$ = -1/m, where, the slope of $\overline{AB}$ = m

Each option is analyzed as follows;

1) A(3, 3) and B(12, 6) C(6, 52)

The slope of $\overline{AB}$ = (6 - 3)/(12 - 3) = 0.$\overline 3$

The slope of $\overline{BC}$ = (52 - 6)/(6 - 12) = -7.$\overline 6$ ≠ 1/(0.$\overline 3$) = 3

∴ $\overline{AB}$ is not perpendicular to $\overline{BC}$ and ∠ABC ≠ 90°

2) A(-10, 5) and B(12, 16) C(16, -6)

The slope of $\overline{AB}$ = (16 - 5)/(12 - (-10)) = 0.5

The slope of $\overline{BC}$ = (-6 - 16)/(16 - 12) = -5.5 ≠ 1/(0.5) = 2

For point C(18, 4), we have;

The slope of $\overline{BC}$ = (4 - 16)/(18 - 12) = -2

∴ A(-10, 5) and B(12, 16) is perpendicular to C(18, 4)

∴ ∠ABC = 90°

3) A(-8, 3) and B(12, 8) C(18, 4)

The slope of $\overline{AB}$ = (8 - 3)/(12 - (-8)) = 0.25

The slope of $\overline{BC}$ = (4 - 8)/(18 - 12) = -0.$\overline 6$ ≠ 1/(0.25) = 4

∴ $\overline{AB}$ is not perpendicular to $\overline{BC}$ and ∠ABC ≠ 90°

4) A(12, -14) and B(-16, 21) C(-11, 25)

The slope of $\overline{AB}$ = (21 - (-14))/((-16) - 12) = -1.25

The slope of $\overline{BC}$ = (25 - (21))/((-11) - (-16)) = 0.8 = 1/(-1.25)

∴ $\overline{AB}$ is perpendicular to $\overline{BC}$ and ∠ABC = 90°

5) A(-12, -19) and B(20, 45)

C(16, -6)

The slope of $\overline{AB}$ = (45 - (-19))/(20 - (-12)) = 2

A(30, 20) and B(-20, -15)

The slope of $\overline{AB}$ = (-15 - 30)/(-20 - 20) = 1.125

The slope of $\overline{BC}$ = (6 - 52)/(12 - 6) = -5.5 ≠ 1/(0.5) = 2

6) The slope of $\overline{BC}$ = (-6 - 16)/(16 - 12) = -5.5 ≠ 1/(0.5) = 2

∴ $\overline{AB}$ is not perpendicular to $\overline{BC}$ and ∠ABC ≠ 90°