If area of a triangle is 35 sq unit with vertices (2,-6), (5,4) and (k,4), then k is
a) 12
b) -2
c) -12, -2
d) 12, -2

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Solution

Given, $\frac{1}{2}$ $∣ ∣ ∣ ∣ \begin{matrix} 2 & - 6 & 1 5 & 4 & 1 k & 4 & 1 \end{matrix} ∣ ∣ ∣ ∣$ =35
i.e. 2(4-4)+6(5-k)+1(20-4k)=70

30-6k +20-4k= 70
On taking positive sign, $- 10 k + 50 = 70 \Rightarrow - 10 k = 20 \Rightarrow k = - 2$
On taking negative sign, $- 10 k + 50 = - 70 \Rightarrow - 10 k = - 120 \Rightarrow k = 12$
$∴ k = 12, - 2$ . Hence, the correct option is (d).

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If area of a triangle is 35 sq unit with vertices (2,-6), (5,4) and (k,4), then k is a) 12 b) -2 c) -12, -2 d) 12, -2

Given, 12∣∣ ∣∣2−61541k41∣∣ ∣∣=35 i.e. 2(4-4)+6(5-k)+1(20-4k)=70 30-6k +20-4k= 70 On taking positive sign, −10k+50=70⇒−10k=20⇒k=−2 On taking negative sign, −10k+50=−70⇒−10k=−120⇒k=12 ∴k=12,−2. Hence, the correct option is (d).

If area of a triangle is 35 sq unit with vertices (2,-6), (5,4) and (k,4), then k is
a) 12
b) -2
c) -12, -2
d) 12, -2