Can you draw a triangle with vertices (1,5) (5,8) and (13,14)
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Answer:
Step-by-step explanation:
Area of a triangle = \frac{1}{2} | (x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) |
= \frac{1}{2} | 1(8-14) + 5(14-5) + 13(5-8) |
= \frac{1}{2} | 1(-6) + 5(9) + 13(-3) |
= \frac{1}{2} | -6 + 45 - 39 |
= \frac{1}{2} | 45 - 45 |
= \frac{1}{2} | 0 |
= \frac{0}{2} = 0
The area of the triangle is equal to zero. So we can not form a triangle.
The given points are collinear
1 and 5 = x1 and y1
5 and 8 =x2and y2
13 and 14 = x3 and y3
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