
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 \[6\,\text{sq.units}\].
Answer
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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 \[A = \dfrac{1}{2}[x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2})]\].
Complete step-by-step answer:
Given the terms \[3k-2, 2k^{2}-5k+8\] and \[4k+3\] are the consecutive terms of an AP.
Using the properties of arithmetic progression, the difference between the consecutive terms are equal.
This implies,
\[2k^{2}-5k+8-(3k-2) = 4k+3-(2k^{2}-5k+8)\]
Solving them as follows:
\[\begin{align*}2k^{2}-5k+8-(3k-2) &= 4k+3-(2k^{2}-5k+8)\\ 2k^{2}-8k+10 &= -2k^{2}+9k-5\\
4k^{2}-17k+15 &= 0\\ 4k^{2}-12k-5k+15 &=0\\ 4k(k-3)+3(k-3) &=0\\ (k-3)(4k+3) &=0\\ k &=3, -\dfrac{3}{4}\end{align*}\]
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.
Note: The area of the triangle can also be evaluated using the lengths of the sides of the triangle, and then using Heron's formula to calculate the area.
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 \[A = \dfrac{1}{2}[x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2})]\].
Complete step-by-step answer:
Given the terms \[3k-2, 2k^{2}-5k+8\] and \[4k+3\] are the consecutive terms of an AP.
Using the properties of arithmetic progression, the difference between the consecutive terms are equal.
This implies,
\[2k^{2}-5k+8-(3k-2) = 4k+3-(2k^{2}-5k+8)\]
Solving them as follows:
\[\begin{align*}2k^{2}-5k+8-(3k-2) &= 4k+3-(2k^{2}-5k+8)\\ 2k^{2}-8k+10 &= -2k^{2}+9k-5\\
4k^{2}-17k+15 &= 0\\ 4k^{2}-12k-5k+15 &=0\\ 4k(k-3)+3(k-3) &=0\\ (k-3)(4k+3) &=0\\ k &=3, -\dfrac{3}{4}\end{align*}\]
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.
Note: The area of the triangle can also be evaluated using the lengths of the sides of the triangle, and then using Heron's formula to calculate the area.
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