Straight Lines - Locus
Triangle ABC with AB = 13, BC = 5, and AC = 12 slides on the coordinates axes with A and B on the
positive x-axis and positive y-axis respectively. The locus of vertex C is a line 12x - ky = 0. Then the value of k
is _
18 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.yo is it 5?
although i wouldnt say my method is proper but still
actually got the ans now
yes it is 5
u can still send ur method im still willing to learn new methods
i took C as (h,k) A as (a,0) B as (0,b) now a^2 + b^2=169 ---1
h^2 + (k-b)^2=25 ---2
k^2 + (a-h)^2=144 ---3
doing 3-2 we get
2(bk-ah) + a^2-b^2-119=0
now since its given its of form y=mx so i did a^2-b^2-119=0 ---4
now we have two eqn to solve for a and b solve it and get locus 12x-5y=0
howd u do it btw
could we have done a^2-b^2-119=0 if we were not given the locus is of form y=mx
i didnt do it. i found a satisfactory solution online
basically
2 main tricks
first
OBCA is a cyclic quadilateral
cuz
sum of opp angles is 180 (BOA & BCA)
and second
if u let theta be the angle made by OC with origin be theta (u would do this naturally cuz u want tan(theta))
ohh right
this is much shorter
now BC is arc
so it will subtend 90 - theta (green on A too)
now tan(90 - t) = 5/12
tan(t) = 12/5
tan(t) is the slope
hmm okay
got it
y = 12/5 * x
god level q tbh
yeahh nice proof
not mine btw
doubtnut
source : cengage
okay np
+solved vj25_
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