Question

Find the equation of the line through (12/5,1), forming with the axes a triangle of area 5.

Answer 1

QUESTION:

Find the equation of the line through (12/5,1), forming with the axes a triangle of area 5.

SOLUTION:

Let's write an equation of straight line in slope-intercept form:

$$y = m \cdot x + b$$

This line crosses y-axis at the point $(0; b)$ and x-axis at the point $(-\frac{b}{m}; 0)$. Hence, it forms with the axes a right triangle. The area of this triangle is $S = \frac{1}{2} \cdot |b| \cdot \left|\frac{b}{m}\right|$. We use here the absolute values of slope and intercept, because they could be negative.

So, as line go through the point $\left(\frac{12}{5}, 1\right)$, and area of triangle is 5, we can write a system of equations:

$$\left\{ \begin{array}{l} m \cdot \frac{12}{5} + b = 1 \\ \frac{1}{2} |b| \cdot \left| \frac{b}{m} \right| = 5 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 12 \cdot m + 5b = 5 \\ |b| \cdot \left| \frac{b}{m} \right| = 10 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 12 \cdot m + 5b = 5 \\ \frac{b^2}{m} = 10 \end{array} \right.$$

Let's consider two cases:

a) $m > 0$, hence $|m| = m$ and:

$$\left\{ \begin{array}{l} 12 \cdot m + 5b = 5 \\ \frac{b^2}{m} = 10 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 12 \cdot m + 5b = 5 \\ m = \frac{b^2}{10} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 12 \cdot \frac{b^2}{10} + 5b = 5 \\ m = \frac{b^2}{10} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} \frac{6}{5} \cdot b^2 + 5b = 5 \\ m = \frac{b^2}{10} \end{array} \right. \Rightarrow$$ $$\Rightarrow \left\{ \begin{array}{l} 6 \cdot b^2 + 25b - 25 = 0 \\ m = \frac{b^2}{10} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} b_1 = \frac{-25 + \sqrt{25^2 - 4 \cdot 6 \cdot (-25)}}{2 \cdot 6} \\ b_2 = \frac{-25 - \sqrt{25^2 - 4 \cdot 6 \cdot (-25)}}{2 \cdot 6} \\ m_1 = \frac{b_1^2}{10} \\ m_2 = \frac{b_1^2}{10} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} b_1 = \frac{-25 + 35}{12} \\ b_2 = \frac{-25 - 35}{12} \\ m_1 = \frac{b_1^2}{10} \\ m_2 = \frac{b_1^2}{10} \end{array} \right.$$ $$\Rightarrow \left\{ \begin{array}{l} b_1 = \frac{5}{6} \\ b_2 = -5 \\ m_1 = \frac{5}{72} \\ m_2 = \frac{5}{2} \end{array} \right.$$

b) $m < 0$ , hence $|m| = -m$ and:

\begin{array}{l} \left\{ \begin{array}{l} 12 \cdot m + 5b = 5 \\ \frac{b^2}{-m} = 10 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 12 \cdot m + 5b = 5 \\ m = \frac{-b^2}{10} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} -12 \cdot \frac{b^2}{10} + 5b = 5 \\ m = \frac{-b^2}{10} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} - \frac{6}{5} \cdot b^2 + 5b = 5 \\ m = \frac{-b^2}{10} \end{array} \right. \\ \Rightarrow \left\{ \begin{array}{l} 6 \cdot b^2 - 25b + 25 = 0 \\ m = \frac{-b^2}{10} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} b_3 = \frac{25 + \sqrt{25^2 - 4 \cdot 6 \cdot 25}}{2 \cdot 6} \\ b_4 = \frac{-25 - \sqrt{25^2 - 4 \cdot 6 \cdot 25}}{2 \cdot 6} \\ m_3 = \frac{-b_3^2}{10} \\ m_4 = \frac{-b_4^2}{10} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} b_3 = \frac{25 + 5}{12} \\ b_4 = \frac{25 - 5}{12} \Rightarrow \\ m_1 = \frac{b_3^2}{10} \\ m_2 = \frac{b_4^2}{10} \end{array} \right. \\ \end{array} \right. \Rightarrow \left\{ \begin{array}{l} b_3 = \frac{5}{2} \\ b_4 = \frac{5}{3} \\ m_3 = -\frac{5}{8} \\ m_4 = -\frac{5}{18} \end{array} \right.

So, we've obtained four straight lines:

$$\begin{array}{l} y = \frac{5}{72} x + \frac{5}{6} \\ y = \frac{5}{2} x - 5 \\ y = -\frac{5}{8} x + \frac{5}{2} \\ y = -\frac{5}{18} x + \frac{5}{3} \\ \end{array}$$

Let's draw the plots:

The plot of $y = \frac{5}{72} x + \frac{5}{6}$

The plot of $y = \frac{5}{2} x - 5$

The plot of $y = -\frac{5}{8} x + \frac{5}{2}$

The plot of $-\frac{5}{18} x + \frac{5}{3}$

ANSWER

$$y = \frac {5}{72} x + \frac {5}{6}$$ $$y = \frac {5}{2} x - 5$$ $$y = - \frac {5}{8} x + \frac {5}{2}$$ $$y = - \frac {5}{18} x + \frac {5}{3}$$