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)

Usingcolor(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 usingcolor(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 usingcolor(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