Question

If area of triangle is 24 sq units with vertices (2,−6),(5,4) and (k,4). Then find the value of k. Solve using determinants.

Answer 1

Given

• Area of triangle is 24 square units with vertices ( 2,-6), ( 5,4) and ( k, 4)

To find

• The value of k

Formula

• Solve using determinants method
• $\tiny\sf \: \: The \: area \: of \: the \: triangle \: is \: given \: by \: \triangle = \frac{1}{2} \begin{gathered}\left[\begin{array}{ccc}\sf x_{1}& \sf \: y_{1} &1 \\ \sf x_{2}& \sf \: y_{2} &1 \\ \sf x_{3}& \sf \: y_{3} &1\end{array}\right]\end{gathered}$

Solution

Given that the area of the triangle is 24 square units and vertices are ( 2,-6), ( 5,4) and ( k, 4), we know the formula

$\purple{\boxed{\tiny\sf \: \: The \: area \: of \: the \: triangle \: is \: given \: by \: \triangle = \frac{1}{2} \begin{gathered}\left[\begin{array}{ccc}\sf x_{1}& \sf \: y_{1} &1 \\ \sf x_{2}& \sf \: y_{2} &1 \\ \sf x_{3}& \sf \: y_{3} &1\end{array}\right]\end{gathered}}}$

Here,

‎ ‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎‎Area of triangle is 24 square units

‎ ‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎‎Since area is always positive,

‎ ‎ So triangle can have positive and negative value

Therefore,

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎‎ Triangle = ± 24 square units,

And also,

$\begin{gathered}\frak{\;value }\begin{cases}\green{\sf x_1 = 2 \: \: \: and \: \: \: y_1 = 6}\\\ \green{ \sf x_2 = 5 \: \: \: and \: \: \: y_2 = 4}\\ \green{ \sf x_3 = k \: \: \: and \: \: y_3 = 4 }\\\end{cases}\end{gathered}$

‎ ‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎‎ Putting value in formula we get,

$\sf \longmapsto \: \: \pm24 = \frac{1}{2} \begin{gathered}\left[\begin{array}{ccc}\sf 2& \sf - 6& \sf1 \\ 5&4&1 \\ \sf \: k&4&1\end{array}\right]\end{gathered}$

$\\ \\ \longmapsto\sf \: \pm24 \bigg \lgroup2 \begin{gathered}\left |\begin{array}{ccc}\sf 4& \sf \: 1 \\ \sf 4& \sf \: 1\\ \end{array}\right |\end{gathered} - ( - 6)\begin{gathered}\left |\begin{array}{ccc}\sf 5& \sf \: 1 \\ \sf k& \sf \: 1\\ \end{array}\right |\end{gathered} + 1\begin{gathered}\left |\begin{array}{ccc}\sf 5& \sf \: 4\\ \sf k& \sf \: 4\\ \end{array}\right |\end{gathered} \bigg \rgroup$

$\\ \\ \longmapsto\sf \: \pm24 = \frac{1}{2} \bigg \lgroup2(4 - 4) + 6(5 - k) + 1(20 - 4k) \bigg \rgroup \\ \\ \\ \longmapsto\sf \: \pm24 = \frac{1}{2} \bigg \lgroup2(0) + 6(5 - k) + (20 - 4k) \bigg \rgroup \\ \\ \\ \longmapsto\sf \: \pm24 = \frac{1}{2} \bigg \lgroup(30 - 6k )+( 20 - 4k )\bigg \rgroup \\ \\ \\ \longmapsto\sf \: \pm24 = \frac{1}{2} \bigg \lgroup(30 - 20) +( 6k - 4k )\bigg \rgroup \\ \\ \longmapsto\sf \: \pm24 = \frac{1}{2} (50 - 10k) \\ \\ \\\longmapsto\sf \: \pm \frac{ 24 \times 2 }{1}= (50 - 10k)\\ \\ \\ \longmapsto\sf \: \pm48 = (50 - 10k) \\ \\ \\ \bf \: so, \: 48 = 50 - 10k \: \: \: or - 48 = 50 - 10k$

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‎ ‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎‎Solving with positive sing

$\bf \: 48 = 50 - 10k \\ \\ \sf \: \longmapsto48 = 50 - 10k \\ \\ \sf \: \longmapsto \: 48 - 50 = 10k \\ \\ \sf \: \longmapsto - 2 = 10k \\ \\ \sf \: \longmapsto10k = - 2 \\ \\ \sf \: \longmapsto k = \cancel \frac{10}{ - 2} \\ \\ \sf \: \longmapsto \: k = - 5$

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‎ ‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎‎Solving with negative sing

$\bf \: - 48 = (50 - 10k) \\ \\ \sf\longmapsto \: - 48 = (50 - 10k) \\ \\ \sf \longmapsto - 48 - 50 = - 10k \\ \\ \sf \longmapsto - 98 = - 10k \\ \\ \sf \longmapsto \: k = \frac{ \cancel- 98}{ \cancel - 10} \\ \\ \sf \longmapsto k=9.8$

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Therefore

‎ ‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎‎‎ ‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎‎ ‎‎ ‎ ‎ ‎‎the value of k is - 5 and 9.8