Question

The line which is concurrent with the
lines 2x + 3y = 7, 2x = 3y + 1 and
passing through the origin is​

Answer 1

$\textbf{Given:}$

$\text{Lines are 2\,x+3\,y-7=0 and 2\,x-3\,y-1=0 }$

$\textbf{To find:}$

$\text{The line which is concurrent with the given lines}$

$\text{and passing through the origin​}$

$\textbf{Solution:}$

$\text{The equation of the line which is concurrent with the given}$

$\text{lines can be written as}$

$(2\,x+3\,y-7)+\lambda(2\,x-3\,y-1)=0$------(1)

$\text{It passes through (0,0)}$

$(2(0)+3(0)-7)+\lambda(2(0)-3(0)-1)=0$

$-7-\lambda=0$

$\implies\bf\lambda=-7$

$\text{Now, (1) becomes}$

$(2\,x+3\,y-7)-7(2\,x-3\,y-1)=0$

$-12x+24\,y=0$

$\implies\boxed{\bf\,x-2\,y=0}$

$\textbf{Answer:}$

$\text{The required line is \bf\,x-2y=0 }$

Answer 2

Answer:

Step-by-step explanation:

The point of intersection of lines,

5x+7y=3...(1)  and  

2x−3y=7...(2)

Solving (1) and (2) , we get,

(x,y)=(2,−1)

Therefore the equation of line passing through (0,0)  and (2,−1)  is  

(y−0)=  

2−0

−1−0

​  

(x−0)

2y=−x

∴x+2y=0  is the required equation