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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Match the question given under Column $C_1$ with their appropriate answer given under the Column $C_2$ for the following : The equation of the line through the intersection of the lines 2x – 3y = 0 and4x – 5y = 2 and

Coulmn $C_1$ Coulmn $C_2$
(a) through the point (2, 1) is (i) 2x – y = 4
(b) perpendicular to the line x + 2y + 1 = 0 is (ii) x + y – 5 = 0
(c) parallel to the line 3x – 4y + 5 = 0 is (iii) x – y –1 = 0
(d) equally inclined to the axes is (iv) 3x – 4y – 1 = 0

 

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Toolbox:
  • Equation of a line passing through the intersection of two lines $a_1x+b_1y+c_1=0$ and $ a_2x+b_2y+c_2=0$ is $(a_1x+b_1y+c_1) + \lambda (a_2x+b_2y+c_2)=0$
  • Slope of a line is $ -\large\frac{coefficient \: of \: x}{coefficient \: of \: y}$
Step 1 :
Let the required equation be
$2x-3y+ \lambda (4x-5y-2)=0$----------(1)
It passes through the point (2,1) is
$2(2)-3(1)+ \lambda (4(2)-5(1)-2)=0$------------(1)
$ \Rightarrow 1+ \lambda = 0$
$ \Rightarrow \lambda = -1$
Hence the equation of the line is
$ 2x-3y+(-1) [ 4x-5y-2]=0$
$ \Rightarrow x-y-1=0$
(a) through the point (2,1) , matches (iii) x-y-1=0
Slope of the line (1)
$x(2+4 \lambda )-y (3+5\lambda )-2\lambda=0$
is $m = \large\frac{-(2+4\lambda )}{-(3+5\lambda)}$$ = \large\frac{2+4\lambda}{3+5 \lambda}$
Slope of the line $x+2y+1=0$ is
$ -\large\frac{1}{2}$
$ \therefore \large\frac{2+4\lambda}{3+5\lambda} $$ \times -\large\frac{1}{2} $$ = -1$
$ \Rightarrow (2+4\lambda)=2(3+5\lambda)$
$ 2+4\lambda=6+10 \lambda$
$ \Rightarrow \lambda = -\large\frac{2}{3}$
Hence equation of this required line is
$x \bigg(2+4 \bigg(-\large\frac{2}{3} \bigg)$$ - y \bigg( 3+5 \bigg( -\large\frac{2}{3} \bigg)$$ - 2 \bigg( -\large\frac{2}{3} \bigg) = $$0$
$ x \bigg( -\large\frac{2}{3} \bigg)-y \bigg( -\large\frac{1}{3} \bigg)+ \large\frac{4}{3}=0$
$ \Rightarrow -2x+y+4=0$
$ \therefore 2x-y=4$
(b) perpendicular to the line x+2y+1=0 matches (i) 2x-y=4
Equation of the given line is $3x-4y+5=0$
Slope of this line is $ - \bigg( \large\frac{3}{-4} \bigg)$
$ = \large\frac{3}{4}$
Since the above line is parallel to the line (1) their slopes are equal.
(i.e) $ \large\frac{2+4\lambda}{3+5\lambda}$$ = \large\frac{3}{4}$
$ \Rightarrow 4(2+ 4\lambda )=3(3+5\lambda)$
$8+16\lambda = 9+ 15\lambda$
$ \Rightarrow \lambda = 1$
Hence the equation of the required line is $x (2+4(1))-y(3+5(+1)-2(1)=0$
(i.e) 6x-8y-2=0$
or $3x-4y-1=0$
(c) parallel to the line 3x-4y+5=0 matches (iv) 3x-4y-1=0
When it is equally inclined between the axes then $m = \pm 1$
When m = -1
$ \therefore \large\frac{2+4\lambda}{3+5\lambda}=-1$
$ \Rightarrow 2+4\lambda = -1(3+5\lambda)$
$ \Rightarrow 2+4\lambda = -3-5\lambda$
$ \Rightarrow -9\lambda = 5$
$ \therefore \lambda = -\large\frac{5}{9}$
Hence the equation of the required line is $x \bigg(2+4 \bigg( -\large\frac{5}{9} \bigg)\bigg)$$ - y\bigg( 3+5 \bigg( -\large\frac{5}{9} \bigg) \bigg) $$ - 2 \bigg( -\large\frac{5}{9} \bigg)=$$0$
(i.e) $x\bigg( -\large\frac{2}{9} \bigg) - y \bigg( \large\frac{2}{9} \bigg) + \large\frac{10}{9}=$$0$
$ \Rightarrow 2x+2y-10=0$
or $x+y-5=0$
(d) equally inclined to the axes is matches (ii) x+y-5=0
answered Jul 11, 2014 by thanvigandhi_1
edited Jul 11, 2014 by thanvigandhi_1
 

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