Question

Find the value of a for which the area of the triangle formed by the points A (a, 2a), B (-2, 6) and C (3, 1) is 10 square units.

Answer 1

Step-by-step explanation:

Hi Mate !!

Given :- area of ∆ABC = 10 unit²

x1 = a

x2 = -2

x3 = 3

y1 = 2a

y2 = 6

y3 = 1

• area of ∆ABC

= \frac{1}{2} (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))

10 = \frac{1}{2} (a(6 - 1) - 2(1 - 2a) + 3(2a - 6)

10 = \frac{1}{2 } (5a - 2 + 2a + 6a - 18)

10 = \frac{1}{2} (13a - 20)

20 = 13a - 20

20 + 20 = 13a

40 = 13a

40/13 = a

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