- #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: