Express y-Coordinate of Point...
Date: 8/9/96 at 9:54:31 From: Anonymous Subject: Express y-Coordinate of Point... P and Q are the points of intersection of the line y/2+x/3 = 1 with the x- and y- axes respectively. The gradient of QR is 1/2 and R is the point whose x-coordinate is 2t, where t is positive. Express the y-coordinate in terms of t and evaluate t given that the area of PQR is 21 units squared. I have tried to get the y-coordinate of R which I found to be "t+2"; therefore R = (t, t+2). I don't know how to find t as I know only the three points and nothing else... no angles or lengths of sides... (expect QP which is 5)...
Date: 9/1/96 at 17:29:35 From: Doctor Jerry Subject: Re: Express y-Coordinate of Point... I'm sure you found that P has coordinates (3,0) and Q has coordinates (0,2). The slope/gradient of the line joining Q and R is 1/2 and the x-coordinate of R is 2t, where t>0. Let the coordinates of R be (2t,w). Use Q and R to calculate the slope of QR. This gives (w-2)/(2t-0) = 1/2. So, as you found, w = t+2. So coordinates of R are (2t,t+2). (You wrote (t, t+2), which was probably a copy error.) One way of looking at triangle PQR is to think about it as a trapezoid minus two right triangles. Let O be the origin point and K be (2t,0), which is directly beneath R. Then area of PQR is area of trapezoid OQRK minus right triangle OPQ minus right triangle PRK. These areas are easy to calculate. Good luck. -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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