Use distance formula to show that the points (cosec^2 theta ,0) , (0,sec^2 theta) and (1,1) are collinear?

1 Answer
Jun 27, 2018

Please see below.

Explanation:

We know the distance formula:

The distance between #color(Brown)(P(x_1,y_1) and Q(x_2,y_2)# is:
#color(Brown)(PQ=sqrt((x_1-x_2)^2+(y_1-y_2)^2)...to(D)#

For simplicity we take ,#(theta to x)#

#A(csc^2x,0) ,B(0,sec^2x) and C(1,1)#

Using#color(brown)( (D) #we get

#(AB)^2=(csc^2x-0)^2+(0-sec^2x)^2#

#color(white)((AB)^2)=(1+cot^2x)^2+(1+tan^2x)^2#

#color(white)((AB)^2)=(1+1/tan^2x)^2+(1+tan^2x)^2#

#color(white)((AB)^2)=(tan^2x+1)^2/tan^4x+(1+tan^2x)^2/1#

#color(white)((AB)^2)=(1+tan^2x)^2[1/tan^4x+1]#

#color(white)((AB)^2)=(sec^2x)^2[(1+tan^4x)/tan^4x]#

#:.(AB)^2=(sec^2x/tan^2x)^2[1+tan^4x]#

#=>color(red)(AB=sec^2x/tan^2xsqrt(1+tan^4x)...to(I)#

Again using#color(brown)( (D) #we get

#(AC)^2=(csc^2x-1)^2+(0-1)^2#

#=>(AC)^2=(cot^2x)^2+1#

#=>(AC)^2=1/tan^4x+1=(1+tan^4x)/tan^4x#

#=>(AC)^2=(1+tan^4x)/(tan^2x)^2#

#=>color(blue)(AC=sqrt(1+tan^4x)/tan^2x...to(II)#

Again using#color(brown)( (D) #we get

#(CB)^2=(0-1)^2+(sec^2x-1)^2=1+(tan^2x)^2#

#=>(CB)^2=1+tan^4x#

#=>color(blue)(CB=sqrt(1+tan^4x)...to(III)#

Adding #(II) and(III)# we get

#AC+CB=sqrt(1+tan^4x)/tan^2x+sqrt(1+tan^4x)/1#

#=>AC+CB=sqrt(1+tan^4x)[1/tan^2x+1]#

#=>AC+CB=sqrt(1+tan^4x)[(1+tan^2x)/tan^2x]#

#=>AC+CB=sqrt(1+tan^4x)[sec^2x/tan^2x] or#

#color(red)(AC+CB=sec^2x/tan^2xsqrt(1+tan^4x)...to(IV)#

From #(I) and (IV)# we can say that

#color(violet)(AC+CB=AB=>A ,B ,C # #"are "color(violet)"collinear points. "#

Note:

#(i)csc^2theta-cot^2theta=1=>csc^2theta-1=cot^2theta#

#(ii)sec^2theta-tan^2theta=1=>sec^2theta-1=tan^2theta#