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winstonbendariovvygn
06.05.2021 •
Mathematics
An ABC triangle with vertices A (2, -3), B (-2,1) and C (0, -3) is given. The bottom of the weight line from vertex A is only:
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Ответ:
A( 0, 2), B (2, 5), and C (−1, 7)
Part AThe length sides of the triangle are
AB , AC and BC
For AB A( 0, 2), B (2, 5)
| AB | = √ ( 2 - 0)² + ( 5 - 2)²
= √ 2² + 3² = √ 4 + 9
= √13 units
For AC A( 0, 2), C (−1, 7)
| AC | = √ ( - 1 - 0)² + (7-2)²
= √1 + 5² = √ 1 + 25
= √ 26 units
For BC B (2, 5), C (−1, 7)
| BC | = √ ( - 1 - 2)² + (7-5)²
= √ (-3)² + 2² = √ 9 + 4
= √13 units
Part B.For AB A( 0, 2), B (2, 5)
Slope = 5 - 2 / 2 - 0 = 3/2
For AC A( 0, 2), C (−1, 7)
Slope = 7-2/ -1-0 = 5/-1 = - 5
For BC B (2, 5), C (−1, 7)
Slope = -1-2/7-5 = -3/2
Part CThe triangle is an isosceles triangle since two sides | AB | and | BC | are equal that's they both have √13units.
Hope this helps you