Elements of analytical geometry and the differential and integral calculus . STRAIGHT LINES. 17 PROPOSITION V. To find the equation of a straight line which shall pass througha given point and make a given angle with a given line. The equation of the given line must be in the form y^ax-\-h. (1) Because the other line must pass through a given point itsequation must be (Prop. III.) y—y^a{x—x). (2) We have now to determine the value of a. When a and a are equal, the two lines must be parallel, andthe inclination of the two lines will be greater or less accordingto the relative values of a and a.
Elements of analytical geometry and the differential and integral calculus . STRAIGHT LINES. 17 PROPOSITION V. To find the equation of a straight line which shall pass througha given point and make a given angle with a given line. The equation of the given line must be in the form y^ax-\-h. (1) Because the other line must pass through a given point itsequation must be (Prop. III.) y—y^a{x—x). (2) We have now to determine the value of a. When a and a are equal, the two lines must be parallel, andthe inclination of the two lines will be greater or less accordingto the relative values of a and a. Let PQhQ the given line (thetangent of its angle with the axisof X equal a) and PR the otherline which shall pass throughthe given point P and make agiven angle QPR. The tangentof the angle PRX=:a. Because PRX=PQR-{-QPR. QPR=^PRX—PQR. Tan. QPR=tan.(PRX—PQR.) As the angle QPR is supposed to be known or given, we mayput m to designate its tangent, and m is a known by trigonometry we have a—a. m=t&n.(PRX—PQR) = }-\-aa (3) Whence a = L— This value of a put in (2) gives g—g=(lL^\(x—x)\ 1—7na / for the equation sought. (4) 18 ANALYTICAL GEOMETRY. Corollary 1. When the given inclination is 90°, m its tangentis infinite, and then «==—-. We decide this in the following manner : An infinite quantity cannot be increased, therefore on thata-^m -L m supposition becomes — or -ma —ma Application^.—To make sure that we comprehend this propo-sition and its resulting equation, we give the following example :The equation of a given line is y=2:c-4-6. Draw another line that will in-tersect this at an angle of 45° andpass through a given point P,whose co-ordinates are x=^, 2/= the line J/IVcorrespondingto the equation y=^x-\-6. Locatethe point jP from its given co-or-dinates. Because the angle of intersectionis to be 45°, m=l, a=2. Substituting these values in (4) we havey~2=~3(x—3^).Or 7/=—3x-{-12i. Constructing the line MB wit
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