- #1

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- Homework Statement
- see attached

- Relevant Equations
- straight lines

Find the problem below;

My approach,

if##x=0##, then ##y=b## and if ##y=0##, then ##x=a##, therefore our co-ordinates are ##(a,0)## and ##(0,b)##. The gradient will give us,

$$-3=\frac {b-0}{0-a}$$

It follows that, ##b=3a##, therefore

$$20=\sqrt {(3a-0)^2+(0-a)^2}$$

$$400=9a^2+a^2$$

##a=2\sqrt 10## and ##b=6\sqrt 10##

Any other approach on this.

My approach,

if##x=0##, then ##y=b## and if ##y=0##, then ##x=a##, therefore our co-ordinates are ##(a,0)## and ##(0,b)##. The gradient will give us,

$$-3=\frac {b-0}{0-a}$$

It follows that, ##b=3a##, therefore

$$20=\sqrt {(3a-0)^2+(0-a)^2}$$

$$400=9a^2+a^2$$

##a=2\sqrt 10## and ##b=6\sqrt 10##

Any other approach on this.

Last edited: