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  • A line through the variable point <img alt="\mathrm{\mathrm{A}(\mathrm{a}+1,2 \mathrm{a})}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cmathrm%7BA%7D%28%5Cmathrm%7Ba%7D

A line through the variable point \mathrm{\mathrm{A}(\mathrm{a}+1,2 \mathrm{a})} meets the lines \mathrm{7 x+y-16=0,5 x-y-8=0, x-5 y+8=0} at \mathrm{B,C,D} respectively. Then \mathrm{AC,BC,AD} are in :

Option: 1

\mathrm{A.P.}


Option: 2

\mathrm{G.P.}


Option: 3

\mathrm{A.G.P.}


Option: 4

\mathrm{H.P.}


Answers (1)

best_answer

The equation of a line passing through \mathrm{\mathrm{A}(\mathrm{a}+1,2 \mathrm{a}) \text { is } \frac{\mathrm{x}-(\mathrm{a}+1)}{\cos \theta}=\frac{\mathrm{y}-2 \mathrm{a}}{\sin \theta}}
The coordinates of any point on this line are given by \mathrm{\frac{\mathrm{x}-(\mathrm{a}+1)}{\cos \theta}=\frac{\mathrm{y}-2 \mathrm{a}}{\sin \theta}=\mathrm{r}}
\mathrm{or, (a+1+r \cos \theta, 2 a+r \sin \theta)\\ Let\ A C=r_1,\ A B=r_2\ and\ A D=r_3.}
Then the coordinates of \mathrm{C} are \mathrm{\left(a+1+r_1 \cos \theta, 2 a+r_1 \sin \theta\right)}
This point lies on the line \mathrm{5x-y-8=0.}
\mathrm{\therefore 5\left(a+1+r_1 \cos \theta\right)-\left(2 a+r_1 \sin \theta\right)-8=0\ \ \ \Rightarrow r_1=\frac{3(1-a)}{5 \cos \theta-\sin \theta}\ .......\left ( i \right )}
The coordinates of \mathrm{B} are given by \mathrm{\frac{\mathrm{x}-(\mathrm{a}+1)}{\cos \theta}=\frac{\mathrm{y}-2 \mathrm{a}}{\sin \theta}=\mathrm{r}_2}
or \mathrm{\left(a+1+r_2 \cos \theta, 2 a+r_2 \sin \theta\right)}
This point lies on \mathrm{7x+y-16=6.}Therefore, \mathrm{7\left(a+1+r_2 \cos \theta\right)+\left(2 a+r_2 \sin \theta\right)-16=0}
\mathrm{\Rightarrow r_2=\frac{9(1-a)}{7 \cos \theta+\sin \theta}}
The coordinates of \mathrm{ C } are given by \mathrm{ \frac{\mathrm{x}-(\mathrm{a}+1)}{\cos \theta}=\frac{\mathrm{y}-2 \mathrm{a}}{\sin \theta}=\mathrm{r}_3 }
or \mathrm{ \left(a+1+r_3\ \cos \theta, 2 a+r_3\ \sin\ \theta\right) }
This point lies on \mathrm{ x-5y+8=0. } Therefore, \mathrm{\left(a+1+r_3 \cos \theta\right)-5\left(2 a+r_3 \sin \theta\right)+8=0}
\mathrm{\Rightarrow r_3=\frac{9(1-a)}{5 \sin \theta-\cos \theta}}
We have to prove that \mathrm{ r_{1},r_{2},r_{3}} are in \mathrm{H.P}. i.e. \mathrm{\frac{2}{r_2}=\frac{1}{r_1}+\frac{1}{r_3}}.
Now, \mathrm{\frac{1}{r_1}+\frac{1}{r_3}=\frac{5 \cos \theta-\sin \theta}{3(1-a)}+\frac{5 \sin \theta-\cos \theta}{9(1-a)} \quad[\text { using (i) and (iii) }]}
\mathrm{=\frac{14 \cos \theta+2 \sin \theta}{9(1-a)} \quad=2 / r_2 \quad[\text { using (ii) }]}
Hence \mathrm{r_{1},r_{2},r_{3}} are in \mathrm{H.P.} 

 

 

 

 

Posted by

vishal kumar

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