slope of AB = p/3
slope of line x+3y = 10 is -1/3
since the lines are perpendicular, p/3 = +3/1
p = 9
equation of line constaining A and B must be
3x - y = k
but (0,0) is on it, so k = 0
from 3x - y = 0 , y = 3x
sub into the given equation:
x + 3(3x) = 10
x = 1, then y = 3
they intersect at C(1,3)
segment AC = √(1 + 9) = √10
segment CB = √( (3-1)^2 + (9-3)^2 ) = √40 = 2√10
so OC : CB = √10 : 2√10 = 1 : 2 or 1/2 : 1
Assuming OC : CB = m : 1 ---> m = 1
so p+4m = 9 + 4(1/2) = 11
note: somebody could interpret CB : OC = m : 1 ---> m = 1
they should make the necessary adjustments.
Suppose the equation x+3y-10=0 represents a straight line which is perpendicular to the line segment joining the points A(0,0) and B(3,p), divides it at C in the ratio of m : 1 where p,m belongs to R. Find the value of p+4m.
2 answers
How to find coordinates of C is not clear to me???
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