Question

If vertices of triangle are (2,4) (5,k) (3,10) and it’s area is 15 so units find the value of k

Answer 1

Answer:

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Answer 2

Answer:

$\red { Value \:of \:k } \green {= -8}$

Step-by-step explanation:

$Let \:A(2,4) = (x_{1} , y_{1}) , \\\:B(5,k) = (x_{2} , y_{2}) ,\:and \:C(3,10) = (x_{3} , y_{3}) \:are \\ vertices\:of \:a \: triangle.$

$Area \:of \: \triangle ABC= 15 \:units \:(given)$

$\implies \frac{1}{2} | x_{1}(y_{2} - y_{3}) +x_{2}(y_{3} - y_{1}) +x_{3}(y_{1} - y_{2}) |= 15$

$\implies |2(k-10)+5(10-4)+3(4-k)| = 30$

$\implies |2k-20+5\times 6+12-3k| = 30$

$\implies | 2k-3k-20+30+12| = 30$

$\implies | -k +22 | = 30$

$\implies -k = 30 - 22$

$\implies k = -8$

Therefore.,

$\red { Value \:of \:k } \green {= -8}$

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