Question 1196068
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Given x=2t, find y in terms of x:<br>
{{{y=5-t=5-(1/2)(2t)=5-(1/2)x}}}<br>
The equation of the line in slope-intercept form is {{{y=(-1/2)x+5}}}.<br>
{{{graph(400,200,-2,12,-1,6,(-1/2)x+5)}}}<br>
(a) The point corresponding to t=0 is (0,5).  From that point to (3,9) the difference in x is 3 and the difference in y is 4, so the distance to (3,9) is 5 (by the 3-4-5 Pythagorean Triple).<br>
(b) To avoid the confusion of the square root, I will find the formula for the square of the distance from (3,9) to a point on the line.<br>
{{{d^2=(3-2t)^2+(9-(5-t))^2=4t^2-12t+9+t^2+8t+16=5t^2-4t+25}}}<br>
(c) Find where the square of the distance is 5^2=25:<br>
{{{5t^2-4t+25=25}}}
{{{5t^2-4t=0}}}
{{{t(5t-4)=0}}}<br>
{{{t=0}}} or {{{t=4/5}}}<br>
We already knew t=0 was one solution; t=4/5 is the other.  The point on the line corresponding to t=4/5 is (2t,5-t) = (8/5,21/5).<br>
CHECK: (3-8/5)^2+(9-21/5)^2 = (7/5)^2+(24/5)^2 = 49/25+576/25 = 625/25 = 25<br>
(d) By symmetry, the point on the line that minimizes the distance from (3,9) is halfway between the two points that are the same distance 5 from the line.  That corresponds to t halfway between 0 and 4/5, or t = 2/5.  That point on the line is (2t,5-t) = (4/5,23/5).<br>