Dear student Coordinates of A=3k-5k+1,5k+1k+1Area of △ABC=2 units⇒123k-5k+15+2+1-2-5k+1k+1+75k+1k+1-5=±2⇒73k-5k+1-2-5k+1k+1+75k+1k+1-35=±4⇒21k-35-2k-2-5k-1+35k+7-35k-35k+1=±4⇒14k-66=±4k+1⇒14k-66=4k+1 and 14k-66=-4k+1⇒14k-4k=70 and 14k+4k=62⇒k=7 or 319 Regards

ques 6

Dear student

C o o r d i n a t e s o f A = (\frac{3 k - 5}{k + 1}, \frac{5 k + 1}{k + 1}) A r e a o f △ A B C = 2 u n i t s \Rightarrow \frac{1}{2} [\frac{3 k - 5}{k + 1} (5 + 2) + 1 (- 2 - \frac{5 k + 1}{k + 1}) + 7 (\frac{5 k + 1}{k + 1} - 5)] = \pm 2 \Rightarrow \frac{7 (3 k - 5)}{k + 1} - 2 - \frac{5 k + 1}{k + 1} + \frac{7 (5 k + 1)}{k + 1} - 35 = \pm 4 \Rightarrow \frac{21 k - 35 - 2 k - 2 - 5 k - 1 + 35 k + 7 - 35 k - 35}{k + 1} = \pm 4 \Rightarrow 14 k - 66 = \pm 4 (k + 1) \Rightarrow 14 k - 66 = 4 (k + 1) a n d 14 k - 66 = - 4 (k + 1) \Rightarrow 14 k - 4 k = 70 a n d 14 k + 4 k = 62 \Rightarrow k = 7 o r \frac{31}{9}

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