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  • Let P be any point on ,Q a point on line 7x + y + 3 = 0 and the line x - y + 1 = 0 is perpen

Let P be any point on x^2+y^2=16 ,Q a point on line 7x + y + 3 = 0 and the line x - y + 1 = 0 is perpendicular bisector of PQ. Then P is

Option: 1

(2,1)

 


Option: 2

(1,2)

 


Option: 3

(1,-2)


Option: 4

None of these


Answers (1)

Let \mathrm{R(a, a+1)}

be ⊥ bisector of PQ. Then equation of PQ is 

\mathrm{\frac{x-\alpha}{\cos \theta}=\frac{y-(\alpha+1)}{\sin \theta}} ------------(1)

Its slope is \tan \theta=-1

\mathrm{\sin \theta=\frac{1}{\sqrt{2}} \cos \theta=-\frac{1}{\sqrt{2}}} 

so (1) becomes.


\mathrm{\frac{\mathrm{x}-\alpha}{-\frac{1}{\sqrt{2}}}=\frac{\mathrm{y}-(\alpha+1)}{\frac{1}{\sqrt{2}}}=\mathrm{r}}  (for point Q)

\mathrm{x=a^{-\frac{r}{\sqrt{2}}}, y=a+1+\frac{r}{\sqrt{2}}}

Putting in straight line 7x + y + 3 = 0


\mathrm{\begin{aligned} & 8 a+4-\frac{6 r}{\sqrt{2}}=0 \\ & \frac{6 r}{\sqrt{2}}=8 a+4 \\ & r=\frac{(8 \alpha+4) \sqrt{2}}{6} \end{aligned}}

For point ‘P’:–

\begin{aligned} & \frac{x-\alpha}{-\frac{1}{\sqrt{2}}}=\frac{y-(\alpha+1)}{\frac{1}{\sqrt{2}}}=-r=-\frac{(8 \alpha+4) \sqrt{2}}{6} \\ & x-\alpha=\frac{8 \alpha+4}{6} \\ & x=\alpha+\frac{8 \alpha+4}{6}=\frac{14 \alpha+4}{6} \\ & y=\alpha+1-\frac{(8 \alpha+4)}{6}=\frac{6 \alpha+6-8 \alpha-4}{6}=\frac{-2 \alpha+2}{6} \end{aligned}

Putting  in circle  x2 + y2  =1 6, we get ;

\mathrm{\begin{aligned} & \left(\frac{14 \alpha+4}{6}\right)^2+\left(\frac{-2 \alpha+2}{6}\right)^2=16 \\ & 200 a^2+104 a-556=0 \\ & \Rightarrow 50 a^2+26 a-139=0 \end{aligned}}

We will get a point which is not  in the  alternatives. 

 

 

 

 

Posted by

Ramraj Saini

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