The co-ordinates of the vertices of Triangle ABC are A (4, 1), B(–3, 2) and C(0, k). Given that the area of Triangle ABC is 12 ${u n i t}^{2}$ . Find the value(s) of k.

5 only

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only

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5 and

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Solution

The correct options are
A
5 only

B
only

Area of ΔABC formed by the given points A(4, 1), B(-3, 2) and C(0, k) is

= |4(2 - k) + (-3) (k - 1) + 0(1 - 2)|

= |8 - 4k - 3k + 3| = (11 - 7k)

But area of ΔABC = 12 ${u n i t}^{2}$ ...............(given)

⇒ |11 - 7k| = 24

± (11 -7k) = 24 ⇒ 11 - 7k = 24 or -(11 - 7k) = 24

-7k = 24 - 11 = 13 ⇒ $\frac{- 13}{7}$

or -(11 - 7k) = 24 ⇒ -11 + 7k = 24

⇒ 7k = 24 + 11 = 35 ⇒ k = $\frac{35}{7}$ = 5

Hence the values of k are : 5, $\frac{- 13}{7}$

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