Aditya Gupta
Last Activity: 5 Years ago
once again, we know that the two point form of a st line is given as:y – y1= [(y2 – y1)/(x2 – x1)]*(x – x1), this line represents the eqn of a line passing through P(x1, y1) and Q(x2, y2).
here the two points are given to be (a cosp, a sinp) and (a cosq, a sinq).
so simply substitute x1= acosp, y1= asinp and x2= acosq, y2= asinq, and simplifying you shall obtain
y(cosp – cosq)= x(sinp – sinq) + a(sinqcosp – sinpcosq)
now, we use some trigono identities:
sinqcosp – sinpcosq= sin(q – p)= 2sin[(p – q)/2]cos[(p – q)/2]
cosp – cosq= – 2sin[(p+q)/2]sin[(p – q)/2]
and sinp – sinq= 2cos[(P + Q)/2]sin[(P - Q)/2]
substitute these values above to obtain
y(-sin[(p+q)/2])= x(cos[(P + Q)/2]) + a(cos[(p – q)/2])
or y(sin[(p+q)/2]) + x(cos[(P + Q)/2]) + a(cos[(p – q)/2])= 0
kindly approve :)
