Question

minimum value of area of the traingle formed by the points 0,0 k,k+1 2k, k-1

Answer 1

9/8

Answer 2

The problem asks for the minimum value of the area of a triangle formed by the points $(0,0)$, $(k, k+1)$, and $(2k, k-1)$.

Let the vertices of the triangle be $A(0,0)$, $B(k, k+1)$, and $C(2k, k-1)$. The formula for the area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by:

$$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

Substitute the coordinates of the given points into the formula: Here, $(x_1, y_1) = (0,0)$, $(x_2, y_2) = (k, k+1)$, and $(x_3, y_3) = (2k, k-1)$.

$$\text{Area} = \frac{1}{2} |0((k+1) - (k-1)) + k((k-1) - 0) + 2k(0 - (k+1))|$$ $$\text{Area} = \frac{1}{2} |0(k+1-k+1) + k(k-1) + 2k(-k-1)|$$ $$\text{Area} = \frac{1}{2} |0(2) + k^2 - k - 2k^2 - 2k|$$ $$\text{Area} = \frac{1}{2} |-k^2 - 3k|$$

Since $|-x| = |x|$, we can write:

$$\text{Area} = \frac{1}{2} |k^2 + 3k|$$

To find the minimum value of the area, we need to find the minimum value of the expression $|k^2 + 3k|$. Let $f(k) = k^2 + 3k$. This is a quadratic function, representing a parabola opening upwards. To find the minimum value of $f(k)$, we can find the vertex of the parabola. The x-coordinate of the vertex for a quadratic $ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$. In this case, $a=1$ and $b=3$, so the vertex occurs at:

$$k = -\frac{3}{2(1)} = -\frac{3}{2}$$

Now, substitute this value of $k$ back into $f(k)$ to find the minimum value of $f(k)$:

$$f\left(-\frac{3}{2}\right) = \left(-\frac{3}{2}\right)^2 + 3\left(-\frac{3}{2}\right)$$ $$f\left(-\frac{3}{2}\right) = \frac{9}{4} - \frac{9}{2}$$

To subtract, find a common denominator:

$$f\left(-\frac{3}{2}\right) = \frac{9}{4} - \frac{18}{4} = -\frac{9}{4}$$

So, the minimum value of $k^2 + 3k$ is $-\frac{9}{4}$.

Now we need to find the minimum value of $|k^2 + 3k|$. Since $k^2 + 3k$ can take negative values, its minimum value is $-\frac{9}{4}$. The value of $|k^2 + 3k|$ will be minimized when $k^2+3k$ is closest to zero. The closest value to zero that $k^2+3k$ takes is its minimum value, $-\frac{9}{4}$. Therefore, the minimum value of $|k^2 + 3k|$ is $\left|-\frac{9}{4}\right| = \frac{9}{4}$.

Finally, substitute this back into the area formula:

$$\text{Minimum Area} = \frac{1}{2} \times \frac{9}{4} = \frac{9}{8}$$

The minimum value of the area of the triangle is $\frac{9}{8}$.