So often we forget the mathematics and calculus we painstakingly learned during engineering. Just for chuckles and grins, here is something different..
The points S, T, U and V have coordinates (s,ms), (t, mt), (u, nu) and (v, nv), respectively.
The lines SV and UT meet the line y = 0 at the points with coordinates (p, 0) and (q, 0), respectively. Show that
p =(m − n)sv/(ms − nv)
,
and write down a similar expression for q.
Given that S and T lie on the circle x^2 + (y − c)^2 = r^2, find a quadratic equation satisfied by s and by t, and hence determine st and s + t in terms of m, c and r.
Given that S, T, U and V lie on the above circle, show that p + q = 0.
And now for something completely different.
Re: And now for something completely different.
wow that's great, remembering old days.
proeng, being not in-touch with advanced mathematics, I am able to solve only the first one. Here we go as;
SV is a line from S (s,ms) to V (v,nv) with some point on this line at (p,0);
So above equation can be derived from two point line equation ((y-y1)/(x-x1)=(y2-y1)/(x2-x1) ; also known as slope equation);
By putting values and rearranging them we can get p =(m − n)sv/(ms − nv) and q = (m - n)tu/(mt - nu).
proeng, being not in-touch with advanced mathematics, I am able to solve only the first one. Here we go as;
SV is a line from S (s,ms) to V (v,nv) with some point on this line at (p,0);
So above equation can be derived from two point line equation ((y-y1)/(x-x1)=(y2-y1)/(x2-x1) ; also known as slope equation);
By putting values and rearranging them we can get p =(m − n)sv/(ms − nv) and q = (m - n)tu/(mt - nu).