Find the value of a for which the area of the triangle formed by the p

1.	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» Given points are A(a,2a), B(−2,6) and C(3,1). It is also said that the area enclosed by them is 10 square units. Area of the triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃) = \(\frac{1}2\) \|x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)\| Given that area of ∆ABC = 10 ∴ 10 = \(\frac{1}2\) \|a(6 – 1) -2 (1 – 2a) + 3(2a - 6)\| ∴ 20 = \|5a – 2 + 4a + 6a - 18\| ∴ 20 = \| 15a – 20\| ∴ 15a – 20 = ± 20 Taking positive sign, 15a – 20 = 20 a = \(\frac{8}3\) Taking negative sign, 15a – 20 = -20 a = 0 Hence, the value of a are 0 and \(\frac{8}3\)

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»

Given points are A(a,2a), B(−2,6) and C(3,1). It is also said that the area enclosed by them is 10 square units.

Area of the triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃)

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

Given that area of ∆ABC = 10

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

∴ 20 = |5a – 2 + 4a + 6a - 18|

∴ 20 = | 15a – 20|

∴ 15a – 20 = ± 20

Taking positive sign,

15a – 20 = 20

a = \(\frac{8}3\)

Taking negative sign,

15a – 20 = -20

a = 0

Hence,

the value of a are 0 and \(\frac{8}3\)

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