Solution 2.2:5d

From Förberedande kurs i matematik 1

Revision as of 09:50, 18 September 2008 by Ian (Talk | contribs)
Jump to: navigation, search

If two non-vertical lines are perpendicular to each other, their gradients \displaystyle k_{1} and \displaystyle k_{2} satisfy the relation

\displaystyle k_{1}k_{2}=-1, and from this we have that the line we are looking for must have a gradient that is given by


\displaystyle k_{2}=-\frac{1}{k_{1}}=-\frac{1}{2}


since the line \displaystyle y=2x+5 has a gradient \displaystyle k_{1}=2 (the coefficient in front of \displaystyle x ).

The line we are looking for can thus be written in the form


\displaystyle y=-\frac{1}{2}x+m


with \displaystyle m as an unknown constant.

Because the point \displaystyle \left( 2 \right.,\left. 4 \right) should lie on the line, \displaystyle \left( 2 \right.,\left. 4 \right) must satisfy the equation of the line,


\displaystyle 4=-\frac{1}{2}\centerdot 2+m


i.e. \displaystyle m=5. The equation of the line is .


\displaystyle y=-\frac{1}{2}x+5