Find the distance of the line passing through the points (a cos alpha, a sin alpha) and
a cos Bita, a sin Bita) from the origin.
Answers
Answer:
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Answer:
the equation of line passing through two points is given by
y - y1 = (y1-y2)*(x-x1)
(x-x1)
this,
y - asinA = ( asinA - asinB) (x - acosA)
(acosA-acosB)
y = (sinA-sinB) x+ asinA - acosA(sinA-sinB)
(cosA-cosB). (cosA-cosB)
y = (sinA-sinB) x/(cosA-cosB)+ a(sinAcosA - sinAcosB - sinAcosA + cosAsinB)/(cosA-cosB)
y = (sinA-sinB)x + asin(B-A)
(cosA-cosB). (cosA-cosB)
m = (sinA - sinB)/(cosA - cosB)
c = asin(B-A)/(cosA-cosB)
distance of line y = mx + c from origin is given by
| c | / √1 + m^2
1+m^2 = 1 + (sinA-sinB)^2/(cosA-cosB)^2
= (cos^2A + cos^2B - 2 cosAcosB + sin^2A + sin^2B - 2sinAsinB)/(cosA-cosB)^2
= 2(1 - cosAcosB - sinAsinB)/(cosA-cosB)^2
= 2(1 - cos(A-B))/(cosA-cosB)^2
= 2*2sin^2(A-B)/2/(cosA-cosB)^2
= 4sin^2(A-B)/2/(cosA-cosB)^2
√(1+m^2) = 2sin(A-B)/2/(cosA-cosB)
|c|/√(1+m^2) = { asin(A-B)/(cosA-cosB)} / 2sin(A-B)/2/(cosA-cosB)
= asin(A-B) / 2sin(A-B)/2
= a*2sin(A-B)/2*cos(A-B)/2 / 2sin(A-B)/2
= a cos(A-B)/2
is the required distance.
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