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

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- 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 $3x+4y+3=0$

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...