A variable straight line throgh the point of intersction of the two lines x/3+y/2=1 and x/2+y/3=1 meets the co-ordinate axes at A and B .find the locus of the middle point of AB?

2 Answers
P dilip_k
Aug 11, 2017

Let the equation of the variable straight line which passes through the point of intersection of the two lines #x/3+y/2=1 and x/2+y/3=1 # be

#(x/3+y/2-1) +lamda(x/2+y/3-1 )=0.....[1]#,

where # lamda# is a parmeter.

Putting #y=0# in [1] we get the intercept from X-axis

#(x/3-1) +lamda(x/2-1 )=0#

#=>(x/3 +lamdax/2)=lambda+1#

#=>x=(6(lambda+1))/(2+3lambda)#

If the variable straight line meets the X- axis at A , then coordinate of A will be

#color(blue)(Ato((6(lambda+1))/(2+3lambda),0) #

Again putting #x=0# in [1] we get the intercept from Y-axis

#(y/2-1) +lamda(y/3-1 )=0#

#=>(y/2 +lamday/3)=lambda+1#

#=>y=(6(lambda+1))/(3+2lambda)#

If the variable straight line meets the Y- axis at B , then coordinate of B will be

#color(blue)(Bto(0,(6(lambda+1))/(3+2lambda)) #

If the mid point of #AB# be #(h,k)#

then
#=>h=(3(lambda+1))/(2+3lambda)#

and

#=>k=(3(lambda+1))/(3+2lambda)#

Now #1/h+1/k=(2+3lambda)/(3(lambda+1))+(3+2lambda)/(3(lambda+1))#

#=>1/h+1/k=(5+5lambda)/(3(lambda+1))#

#=>1/h+1/k=(5(1+lambda))/(3(lambda+1))=5/3#

So putting #h=x and k=y # we get the equation of locus of the mid point of #AB# as

#color(red)(1/x+1/y=5/3)#

Ratnaker Mehta
Aug 13, 2017

# 3(1/x+1/y)=5.#

Explanation:

Let #P(h,k)# be the Mid-pt. of the Sgmnt. #AB,# where, we

take, the pts. #A, and, B# on the X, and, Y-Axis, resp.

We are reqd. to determine the Locus of the pt. #P(h,k).#

In other words, we have to find an Algebraic Eqn. that relates

#h and k.#

Clearly, #A=A(2h,0), B=B(0,2k),# and hence, the eqn. of line

#AB," say "l,# is, #x/(2h)+y/(2k)=1.#

Naming #l_1 : x/3+y/2=1, and, l_2 : x/2+y/3=1,# we recall that,

# l : kx+hy-2hk=0,#

# l_1 :2x+3y-6=0,# and,

# l_2 : 3x+2y-6=0,# are concurrent, as they all pass through #P.#

Evidently, #|(k,h,-2hk),(2,3,-6),(3,2,-6)|=0.#

#:. -2|(k,h,hk),(2,3,3),(3,2,3)|=0.#

#because -2ne0, :. k(9-6)-h(6-9)+hk(4-9)=0.#

#:. 3k+3h=5hk, or, 3(1/h+1/k)=5.#

Converting #(h,k) to"the conventional "(x,y),# we get the desired

locus as, # 3(1/x+1/y)=5.#

Enjoy Maths!

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