John K. answered • 07/25/15
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Math and Engineering Tutor, Professional Engineer
Eq of line is y = (y2 - y1)/(x2 - x1) x + b that it can be solved for any two points (x1,y1),(x2,y2). Then
AB -> y = -2x + 18. Also y = mx + b with m = (y2 - y1)/(x2 - x1). Two lines are perpendicuar if m1*m2 = -1 or the product of their slopes is -1. A line perpendicuar to y = 4x + 5 is y = -1/4x + b. Using C then this line is y=-1/4x+10.5.
Then the two equations y=-1/4+10.5 and y=-2x+18 can be solved for the point P(x,y). Then using distance formulae the ratio AP:PB can be determined with a little calculation.
John K.
A line can be expressed as y=(y2-y1)/x2-x1) + b or y=mx+b with slope m. We are given points A,B, then the line through A,B is y = -2x + 21. We know that the slope of line perpendicular to y=4x+5 is -1/4 as the products of the slopes of perpendicular lines equals -1. Then from the fact that is passes through C we can use slope intercept form to obtain y=-1/4x+10.5 for the perpendicular line through C. Point P is the intersection of these two lines found by solving the two line equations in two unknowns as P = P(6,9). It is probably to best to plot these points on a graph to get a feel for computing distances between points using right triangles or pagathorius theorem. Using this the distance AP is the sqrt(4 + 16) and PB is sqrt(9+36) or AP:PB = sqrt(20)/sqrt(45) = 2/3.
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07/28/15
David K.
07/26/15