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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Find the equation of the line passing through the point of intersection of 2x + y = 5 and x + 3y + 8 = 0 and parallel to the line 3x + 4y = 7.

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

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Toolbox:
  • If two lines are parallel, then their slopes are equal.
  • Equation of a line passing through the point $(x_1,y_1)$ and slope $m $ is $y-y_1=m(x-x_1)$
Step 1:
The point of intersection of the lines can be obtained by solving the two given lines.
$2x+y=5$ and $x+3y+8=0$
$ \qquad 2x+y=5$
$( \times 2 ) x+3y=-8$
____________________
$ \qquad 2x+y=5$
$\qquad 2x+6y=-16$
$ \qquad (-) \quad (-) \quad (+)$
____________________
$ \quad \qquad -5y=21$
$ \Rightarrow y = -\large\frac{21}{5}$$ \therefore x = \large\frac{23}{5}$
Hence the point of intersection is $ \bigg( \large\frac{23}{5}$$, \large\frac{-21}{5} \bigg)$
Step 2 :
It is given that the required line is parallel to the line $3x+4y=7$
$ \therefore $ slope $m = -\large\frac{3}{4}$
$ \therefore $ The equation of the given line is
$ (y-y_1) = m (x-x_1)$
(i.e) $y-\bigg(\large\frac{-21}{5} \bigg)$$=-\large\frac{3}{4} \bigg( x - \large\frac{23}{5} \bigg)$
$y+\large\frac{21}{5}$$ = \large\frac{-3x}{4}+\large\frac{69}{20}$
$ \Rightarrow \large\frac{5y+21}{5}$$= \large\frac{-15x+69}{20}$
$ \Rightarrow 5y+21=\large\frac{-15x+69}{4}$
(i.e) $4(5y+21)=-15x+69$
$ \Rightarrow 20y+84=-15x+69$
$ \Rightarrow 15x+20y=-15$
(i.e) $3x+4y=-3$
Hence the equation of the required line is $3x+4y+3=0$
answered Jul 1, 2014 by thanvigandhi_1
edited Jul 1, 2014 by thanvigandhi_1
 

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