8.x3y¬63y¬x6y¬13x2x3y¬143y¬x14y¬13x134First, write an equation of a line pperpendicularto the given lines. The slope of pis the opposite reciprocal of 13, or 3. Use the y-intercept of y13x2, (0, 2), as one of the endpoints of theperpendicular segment.yy1¬m(xx1)y2¬3(x0)y2¬3x y¬3x2Next, use a system of equations to determine thepoint of intersection of y13x134and p.13x134¬3x213x3x¬2134130x¬230x¬2y¬3(2)2y¬4The point of intersection is (2,4).Then, use the Distance Formula to determine thedistance between (0, 2) and (2,4).d¬(x2x1)2(y2y1)2¬(20)2(42)2¬40¬210The distance between the lines is 2 10, orapproximately 6.32 units.9.1.Graph y34x14and point P. Place thecompass at point P. Make the setting wideenough so that when an arc is drawn, itintersects y34x14in two places. Labelthese points of intersection Aand B.2.Put the compass at point Aand draw an arcbelow the line.3.Using the same compass setting, put the
This is the end of the preview.
access the rest of the document.