Hi everyone. New guy here, been stuck with a locus problem since two days ago, so out of despair i will post it here. a=(5,-3), b=(1,4). Find the equation of the locus of the point c such that area of triangleabc=10. Thank you very much in advance.
Locus: Given pts a=(5,-3), b=(1,4), find locus of c such that area of triangle abc=10
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Can you calculate length of straight line from a to b ?
Steven G
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Can you please explain what the equation of the locus of the point c is??
pka
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Do yo know that the distance from the point \(\displaystyle (h,k)\) to the line \(\displaystyle ax+by+c=0\) is \(\displaystyle \dfrac{|a\cdot h+ b\cdot k +c|}{\sqrt{a^2+b^2}}~?\)
If you do How do you finish?
Hi Denis. So, d square (distance from a to b squared)= (5-1)squared+(4+3) squared. Result is 65. So distance from a to b is the square root of 65. Then the only thing i can think of is that the distance from c to the line ab is equal to 2*10/square root of 65. I get lost at that point.
Hi pka. I have a vague recollection of this formula but i dont really understand it. I am guessing ax+by+c=0 to be the general formula of the equation of the parabola, not the a and b as points mentioned in the problem i posted originally. It sounds too advanced for the book i am reviewing. Doesnt go into parabolas or cones or circles at all. Just gives us a small intro into the locus. There is a solved problem dealing with the midpoint of a straight line, putting the two halves of the line on both sides of the = sign, thus getting rid of the x squared and y squared, and leaving an x+y=4, then it says that the locus of point p is the perpendicular bisector of the line [ab], then it gives 4 problems to solve. The third one is the one i posted originally and i cannot connect the very simple example of the book with this problem i am stuck with.
You need to have a serious talk with your geometry teacher.
Steven G
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I thought that was you.
pka
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The equation of the line \(\displaystyle \ell: \overline{AB}\) is \(\displaystyle 7x+4y-23=0\).
Now if the point \(\displaystyle C ; (p,q)\) the distance \(\displaystyle d(\ell,\overline{AB})\) from \(\displaystyle \ell \text{ to }\overline{AB}\) is \(\displaystyle \dfrac{|7p-4q-23|}{\sqrt{(7)^2+(-4)^2}}\)
The area of \(\displaystyle \Delta ABC\) is one-half length of the base, \(\displaystyle \overline{AB}\), times height \(\displaystyle d(\ell,\overline{AB})\).