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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
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Associated Topics:
High School Coordinate Plane Geometry
High School Geometry

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