On a coordinate plane, triangle A B C is shown. Point A is at (negative 1, 1), point B is at (3, 2), and points C is at (negative 1, negative 1) If line segment BC is considered the base of triangle ABC, what is the corresponding height of the triangle? 0.625 units 0.8 units 1.25 units 1.6 units

1.6 units

Step-by-step explanation:

A(-1, 1)

B(3,2)

C(-1, -1)

BC slope: (2 - (-1))/(3 - (-1)) = 3/4

Equation of a line that pass through A(-1, 1) with slope -4/3 (so that it is perpendicular to segment BC)

y - y1 = m(x - x1)

y - 1 = -4/3(x + 1)

Equation of the line that passes through segment BC:

y - y1 = m(x - x1)

y - 2 = 3/4(x - 3)

These two lines intersect is found solving the next system of equations:

y - 1 = -4/3(x + 1) (eq. 1)

y - 2 = 3/4(x - 3) (eq. 2)

Subtractin eq. 2 to eq. 1, we get :

-1 + 2 = -4/3(x + 1) - 3/4(x - 3)

1 = -4/3x - 4/3 - 3/4x + 9/4

1 + 4/3 - 9/4 = -25/12x

1/12*(-12/25) = x

-0.04 = x

Replacing into eq. 1:

y - 1 = -4/3(-0.04 + 1)

y = -1.28 + 1

y = -0.28

The height of the triangle is the distance between A(-1, 1) and (-0.04, -0.28)

d = √[(-1 - (-0.04))² + (1 - (-0.28))²] = 1.6 units

answer: f true

step-by-step explanation:

ok then what happened.