Straight line in mathematics

Question

A line having slope $-1/5$ meets the $xy$-axes at points $A$ and $B$. If $\text{ar}(\triangle OAB)$ is $10$ where $O$ is the origin, find an equation of the line.

Answer 1

The equation of the line can be expressed as $$y=-\frac{1}{5}x+b.$$ Plug $x=0$ to obtain the intersection with the $y$-axis: it is $b$ and plug $y=0$ to obtain the intersection with the $x$-axis: it is $5b$. Then the area of triangle is given by $$A=\frac{b\cdot5b}2=\frac{5b^2}{2}$$ Solve $$5b^2/2=10\iff b=\pm2$$ The line is $$y=-\frac{1}{5}x\pm2$$

Answer 2

if $$y=-\frac{1}{5}x+n$$ then we get $$A(0;n)$$ and $$B(5n;0)$$ then the area is given by $$S=\frac{1}{2}5n^2$$

Answer 3

Suppose $A (a,0) $ and $B (0,b) $.

The equation if the line $AB $ has the form

$$y=\frac {-1}{5}(x-a) $$ with $$b=\frac {-1}{5}(0-a)=\frac {a}{5} $$

the area of the triangle $\triangle{OAB} $ is

$$\frac {ab}{2}=\frac {a^2}{5}=10$$

then $$ a=\pm 5\sqrt {2} $$ and the equation is $$y=\frac{-1}{5}(x\pm 5\sqrt {2}).$$ $$=\frac {-x}{5}\pm\sqrt {2} $$