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  • A line is drawn through a variable point so as to meet the

A line is drawn through a variable point \mathrm{A(t + 1, 2t)} so as to meet the lines \mathrm{7x + y - 16 = 0, 5 x-y-8=0}, and 
\mathrm{x - 5y + 8 = 0} at B, C and D respectively, then AC, AB, AD are in:

Option: 1

A.P


Option: 2

G.P


Option: 3

H.P


Option: 4

A.C.


Answers (1)

best_answer

Let, \mathrm{A B=r_1, A C=r_2, A D=r_3}

Now any line through \mathrm{\mathrm{A}(\mathrm{t}+1,2 \mathrm{t})} is 

\mathrm{\frac{x-(t+1)}{\cos \theta}=\frac{y-2 t}{\sin \theta}=r_1, r_2, r_3}

Clearly \mathrm{\left[r_1 \cos \theta+(t+1), r_1 \sin \theta+2 t\right]} is point B which lies on \mathrm{7 x+y-16=0} , we get 

\mathrm{7\left\{r_1 \cos \theta+(t+1)+r_1 \sin \theta+2 t-16=0\right.}

\mathrm{\Rightarrow r_1=\frac{9(1-t)}{7 \cos \theta+\sin \theta}}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ....(1)

similarly \mathrm{r_2=\frac{3(1-t)}{5 \cos \theta-\sin \theta}}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ....(2)

and \mathrm{r_3=\frac{9(1-t)}{5 \sin \theta-\cos \theta}}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ....(3)

Now, \mathrm{\frac{1}{r_2}+\frac{1}{r_3}=\frac{5 \cos \theta-\sin \theta}{3(1-t)}+\frac{5 \sin \theta-\cos \theta}{9(1-t)}}

\mathrm{=\frac{14 \cos \theta+2 \sin \theta}{9(1-t)}=\frac{2(7 \cos \theta+\sin \theta)}{9(1-t)}=\frac{2}{r_1}}

\Rightarrow \mathrm{AC, AB, AD \: are\: in\: H.P.}

Posted by

Ritika Kankaria

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