VII

3 ) = ( − 3 , − 5 ) .

It is easy to see that $(x,y)-(u,v)=(x,y)+(-u,-v).$ Also $(x,y)-(x,y)=(0,0).$ Hence $(0,0)$ is to be looked on as the zero ordered couple. For example $(x,y)+(0,0)=(x,y).$

The pictorial representation of the addition of ordered couples is surprisingly easy. 10

plus 0.75em minus 0.25em Let $OP$ represent $(x,y)$ so that $OM=x$ and $PM=y$; let $OQ$ represent $(x_1,y_1)$ so that $O{M}_1=x_1$ and $Q{M}_1=y_1$. Complete the parallelogram $OPRQ$ by the dotted lines $PR$ and $QR$, then the diagonal $OR$ is the ordered couple $(x+x_1,y+y_1)$. For draw $PS$ parallel

to $OX$; then evidently the triangles $OQ{M}_1$ and $PRS$ are in all respects equal. Hence $M{M}^{\prime }=PS=x_1$, and $RS=Q{M}_1$ and therefore

$$\begin{gathered} O{M}^{\prime }=OM+M{M}^{\prime }=x+x_1, \\ R{M}^{\prime }=S{M}^{\prime }+RS=y+y_1. \end{gathered}$$

Thus $OR$ represents the ordered couple as required. This figure can also be drawn with $OP$ and $OQ$ in other quadrants.

It is at once obvious that we have here come back to the parallelogram law, which

was mentioned in VI., on the laws of motion, as applying to velocities and forces. It will be remembered that, if $OP$ and $OQ$ represent two velocities, a particle is said to be moving with a velocity equal to the two velocities added together if it be moving with the velocity $OR$. In other words $OR$ is said to be the resultant of the two velocities $OP$ and $OQ$. Again forces acting at a point of a body can be represented by lines just as velocities can be; and the same parallelogram law holds, namely, that the resultant of the two forces $OP$ and $OQ$ is the force represented by the diagonal $OR$. It follows that we can look on an ordered couple as representing a velocity or a force, and the rule which we have just given for the addition of ordered couples then represents the fundamental laws of mechanics for the addition of forces and