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, negativ

Question

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

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niczorrrr 3 years 2021-08-24T23:57:45+00:00 2 Answers 7 views 0

Answers ( )

  1. taiduc
    0
    2021-08-24T23:58:46+00:00
    Reply

    Answer:

    D. ( LAST ONE ) 1.6 UNITS

    Step-by-step explanation:

    IF (x₁,y₁) ,(x₂,y₂) AND (x₃,y₃) ARE THE VERTICES OF THE TRIANGLE.

    THEN THE AREA OF THE TRIANGLE IS

    \frac{1}2} [X1(y2-y3)+x2(y3-y1)+x3(y1-y2)}

    HERE x₁= -1,y₁=1 ,x₂=3,y₂=2, and x₃= -1,y₃=-1

    THEN THE AREA OF THE TRIANGLE IS

    |\frac{1}{2} {-1(2-(-1))+3(-1-1)+(-1)(1-2)}|\\=|\frac{1}{2} x (-8)|

    =4 square units

    BC is base of the triangle.

    Then the length of BC is = \sqrt{(-1-3)^{2} +(-1-2)^{2}

    = 5 UNITS

    Let the height of the triangle be x.

    We know that

    The area of triangle is

    1/2 x base x height

    = (1/2x5xX) square units

    According to the problem,

    1/2 x 5 x X=4

    –  X =  4×2/5

    X=1.6 UNITS

    Therefore the height of the triangle is 1.6 units.

  2. huyenthanh
    0
    2021-08-24T23:59:21+00:00
    Reply

    Answer:

    D. 1.6

    Step-by-step explanation:

    magic

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