Elements of Analytical Geometry and of The Differential and Integral Calculus

ANALYTICAL GEOMETRY.

SECTION I.
APPLICATION OF ALGEBRA TO GEOMETRY

ARTICLE 1. The relations of Geometrical magnitudes may be expressed by means of algebraic symbols, and the demonstrations of Geometrical theorems may thus be exhibited more concisely than is possible in ordinary language. Indeed, so great is the advantage in the use of algebraic symbols, that they are now employed to some extent in all treatises on Geometry.

(2.) The algebraic notation may be employed with even greater advantage in the solution of Geometrical problems. For this purpose we first draw a fgure which represents all the parts of the problem, both those which are given and those which are required to be found. The usual symbols or letters for known and unknown quantities are employed to denote both the known and unknown parts of the figurc, or as many of them as muy be necessary. We then observe the relations which the scveral parts of the figure bear to each other from which, by the aid of the proper theorems in Geometry we derive as many independent equations as there are unknown quantities employed. The solution of these equations by the ordinary rules of algebra will determine the value of the unknown quantities. This method will be illustrated by a few examples.

Ex.1. In a right-angled triangle, having given the base and sum of the hypothenuse and perpendicular, to find the perpendicular.

Let ABC represent the proposed triangle, right-angled at B. Represent the base AB by [math]\displaystyle{ b }[/math], the perpendicular BC by [math]\displaystyle{ z }[/math], and the sum of the hypothenuse and perpendicular by [math]\displaystyle{ s }[/math]; then the hypothenuse will be represented by [math]\displaystyle{ s-x }[/math]. Then, by Geom., Prop. 11, B. IV.,

[math]\displaystyle{ \mathrm{\overline{AB}^2+\overline{BC}^2=\overline{AC}^2} }[/math];

that is,[math]\displaystyle{ b^2+x^2=(s-x)^2=s^2-2sx+x^2 }[/math].

Taking away [math]\displaystyle{ x^3 }[/math] from each side of the equatuon, we have

[math]\displaystyle{ b^2=s^2-2sx }[/math],

or[math]\displaystyle{ 2sx=s^2-b^2 }[/math];

Whence[math]\displaystyle{ x=\frac{s^2-b^2}{2s} }[/math],

from which we see that in any right-angled triangle, the perpendicular is equal to the square of the sum of the hypothenuse and perpendicular, diminished by the square of the base, and divided by twice the sum of the hypothenuse and perpendicular. Thus, if the base is 3 feet, and the sum of the hypothenuse and perpendicular 9 feet, the expression [math]\displaystyle{ \frac{s^2-b^2}{2s} }[/math] becomes [math]\displaystyle{ \frac{9^2-3^2}{2\times9}=4 }[/math], the perpendicular.