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Determine $k$ so that $3k-2, 2k^{2}-5k+8$ and $4k+3$ are the consecutive terms of an AP? For what value of $k (k>0)$, the area of triangle with vertices $(k,2)$, $(3k,2)$ and $(2,5)$ is $6\,\text{sq.units}$.

Last updated date: 20th Jun 2024
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Hint: If $a$, $b$ and $c$ are the consecutive terms of an AP, then using the properties of arithmetic progression, $b-a = c-b$.
Also, the area of the triangle with the vertices $(x_{1},y_{1})$, $(x_{2},y_{2})$ and $(x_{3},y_{3})$ is given by the formula $A = \dfrac{1}{2}[x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2})]$.

Given the terms $3k-2, 2k^{2}-5k+8$ and $4k+3$ are the consecutive terms of an AP.
$2k^{2}-5k+8-(3k-2) = 4k+3-(2k^{2}-5k+8)$
\begin{align*}2k^{2}-5k+8-(3k-2) &= 4k+3-(2k^{2}-5k+8)\\ 2k^{2}-8k+10 &= -2k^{2}+9k-5\\ 4k^{2}-17k+15 &= 0\\ 4k^{2}-12k-5k+15 &=0\\ 4k(k-3)+3(k-3) &=0\\ (k-3)(4k+3) &=0\\ k &=3, -\dfrac{3}{4}\end{align*}
So, the value of $k = 3, -\dfrac{3}{4}$.
Now, the vertices of the triangle are $(k,2)$, $(3k,2)$ and $(2,5)$, and its area is $6\,\text{sq.u}$.
So, substituting the values into the formula for the area of the triangle, $A = \dfrac{1}{2}[x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2})]$, it gives,
\begin{align*}6 &= \dfrac{1}{2}[k(2-5)+3k(5-2)+2(2-2)]\\ 12 &= -3k+9k\\ 12 &= 6k\\ k&= 2\end{align*}
Therefore, the value of $k$ is 2.