Question

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 1

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.