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

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:

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.