logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XI  >>  Math  >>  Straight Lines
0 votes

Find the equations of the lines through the point of intersection of the lines $x-y+1=0 $ and $ 2x-3y+5 = 0$ and whose distance from the point (3, 2) is $ \large\frac{7}{5}$

$\begin {array} {1 1} (A)\;4x+3y-1=0 ; 3x+4y+6=0 & \quad (B)\;4x-3y+1=0 ; 3x-4y+6=0 \\ (C)\;4x-3y-1=0 ; 3x-4y-6=0 & \quad (D)\;4x+3y-1=0 ; 3x+4y-6=0 \end {array}$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Equation of a line passing through the intersection of two lines is $a_1x+b_1y+c_1 + \lambda (a_2x+b_2y+c_2)=0$
  • Distance of a line $ax+by+c=0$ from a point $(x_1, y_1)$ is $d = \bigg| \large\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}} \bigg|$
Step 1 :
Let the equation of the line passing through the point of intersection be
$(x-y+1)+ \lambda (2x-3y+5)=0$
( i.e) $x(1+2\lambda )-y(1+3\lambda)+1+5\lambda=0$
Step 2 :
Its distance from the point (3,2) is given as $ \large\frac{7}{5}$
$ \therefore \bigg| \large\frac{3(1+2\lambda )-2(1+3\lambda)+1+5\lambda}{\sqrt{(1+2\lambda)^2+(1+3\lambda)^2}} \bigg|$$= \large\frac{7}{5}$
$ \Rightarrow \bigg| \large\frac{3+6\lambda - 2 - 6 \lambda + 1 + 5 \lambda}{\sqrt{1+4\lambda^2+4\lambda+1+6\lambda+9\lambda^2}} \bigg|$$= \large\frac{7}{5}$
$ \Rightarrow \bigg| \large\frac{2+5\lambda}{\sqrt{2+10\lambda+13\lambda^2}} \bigg|$$ = \large\frac{7}{5}$
Step 3 :
Squaring on both sides we get,
$ \large\frac{(2+5\lambda)^2}{13\lambda^2+10\lambda+2}$$=\large\frac{49}{25}$
$ \large\frac{4+20\lambda+25\lambda^2}{13\lambda^2+10\lambda+2}$$=\large\frac{49}{25}$
$25(4+20\lambda+25\lambda^2)=49(13\lambda^2+10\lambda+2)$
$100+500\lambda+625\lambda^2=637\lambda^2+490\lambda+98$
$ \Rightarrow 12\lambda^2-10\lambda-2=0$
$ \Rightarrow 6\lambda^2-5\lambda-1=0$
On factorizing we get,
$(6\lambda+1)(\lambda-1)=0$
Hence $\lambda=-\large\frac{1}{6}$ or $ \lambda=1$
<< Enter Text >>
Step 4 :
The equation of the required line is when $ \lambda=-\large\frac{1}{6}$
$x\bigg(1+2\bigg(-\large\frac{1}{6} \bigg) \bigg) -y \bigg( 1 + 3\bigg(-\large\frac{1}{6} \bigg)\bigg)$$+1+5\bigg(-\large\frac{1}{6} \bigg)=0$
(i.e) $x\bigg(\large\frac{2}{3} \bigg)$$-y\bigg(\large\frac{1}{2} \bigg)$$+\large\frac{1}{6}$$=0$ (i.e) $4x-3y+1=0$
Case (ii)
when $ \lambda = 1$
$x(1+2(1))-y(1+3(1))+1+5(1)=0$
$x(3)-y(4)+6=0$
$3x-4y+6=0$
Hence the equations are
$ 4x-3y+1=0$ and $ 3x-4y+6=0$
answered Jul 2, 2014 by thanvigandhi_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...