Elements of analytical geometry and the differential and integral calculus . o-41^ then ABMN ^Y\\\ be a parallelo-gram. Put AP=x. PM=y. The tan-gent of the angle NAP=a. Thenwill KP=ax. To each of these equals add NM=^b, then we shall hare y=:zax-\-b for the algebraic expression corresponding to the point M, andas J!f is any variable point on the line ML corresponding to thevariations of x, this equation is said to be the equation of theline ML. When b is minus the line is then QL, and cuts the axis YFin D, a point as far beloAv ^ as ^ is above A. Hence we perceive that the equation may represe
Elements of analytical geometry and the differential and integral calculus . o-41^ then ABMN ^Y\\\ be a parallelo-gram. Put AP=x. PM=y. The tan-gent of the angle NAP=a. Thenwill KP=ax. To each of these equals add NM=^b, then we shall hare y=:zax-\-b for the algebraic expression corresponding to the point M, andas J!f is any variable point on the line ML corresponding to thevariations of x, this equation is said to be the equation of theline ML. When b is minus the line is then QL, and cuts the axis YFin D, a point as far beloAv ^ as ^ is above A. Hence we perceive that the equation may represent the equation of any line in the plane TAX. If we give to «, x, and b, their proper signs, in each case ofapplication we may write . y=?c-\-b for the equation of any straight line i7i a plane. To fix in the minds of learners a complete comprehension ofthe equation of a straight line, we give the following practical EXAMPLES. 1. Draw the line whose equation is y=2x-\-S. (1)Then draw the line represented by y=—rr-|-2 (2) and determine where these two lines STRAIGHT LINES. 13 Draw YY and XX at rightangles, and take any conveni-ent space for the unit of mea-sure, as 1, 2, 3, (fee. Equation (1) is true for allv^alues of x and y. It is truethen when a:=0. But whena;=0 the point on the line mustbe on the axis YY, When ir=0. y=3. This shows that the line sought for must cut YY at the point +3. The equation is equally true when y=0. But when y=0,the corresponding point on the line sought must be on theaxis XXy and on the same supposition the equation becomes 0=237+3, Or x=—\\. That is, midway between —1 and —2 is another point in theline which is represented by y=2ar-(-3, but two points in anyright line must define the line : therefore ML is the line sought. Taking equation (2) and making a;=0 will give 2/=2, andmaking 2/=0 will give a;=2 : therefore MQ must be the linewhose equation is y=—a:-|-2, and these two lines with the axisXXform the triangle LMQ, whose b
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