Schaum's Outline of Theory and Problems of Beginning Calculus

32 STRAIGHT LINES [CHAP. 4Fig. 4-12negative.) Let Yl intersect Y2 at point P (see Fig. 4-13). Let LY3 be the line through P and perpendicular toY1. If m3 is the slope of Y3, then, by what we have just shown,m1m3 = - 1 = m1m2 orm3 = rn,Since Lf2 and Y3 pass through P and have the same slope, they must coincide. Since LYl is perpendicularto Y3, Yl is perpendicular to Y2.4.6 Show that the line y = x makes an angle of 45" with the positive x-axis (that is, KPOQ in Fig.4-14 contains 45").Let P be the point on the line y = x with coordinates (1, 1). Drop a perpendicular PQ to the positivex-axis. Then PQ = 1 and ?@ = 1. Hence, XOPQ = XQOP, since they are base angles of isosceles triangleQPO. Since 3c OQP contains 90",XOPQ + XQOP = 180" - XOQP 180" - 90" = 90"Since SOPQ = < QOP, they each contain 45".4.7 Sketch the graph of the equation I x I + I y I = 1.Consider each quadrant separately. In the first quadrant, Ix I = x and Iyl = y. Hence, the equationbecomes x + y = 1; that is, y = -x + 1. This yields the line with slope - 1 and y-intercept 1. This lineintersects the x-axis at (1,O). Hence, in the first quadrant our graph consists of the line segment connecting(1, 0) and (0, 1) (see Fig. 4-15). In the second quadrant, where x is negative and y is positive, 1x1 = -xand I y( = y, and our equation becomes -x + y = 1, or y = x + 1. This yields the line with slope 1 andt't'(b)Fig. 4-13 Fig. 4-14