Question

Given the four lines with equations $x+2y=3,3x+4y=7,2x+3y=4$ and $4x+5y=6$, then these lines are

• A. Concurrent
• B. Perpendicular
• C. The sides of a rectangle
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option:

Step 1. Given equations are as follows

$$\begin{gathered} x+2y=3...\left(i\right) \\ 3x+4y=7...\left(ii\right) \\ 2x+3y=4...\left(iii\right) \\ 4x+5y=6...\left(iv\right) \end{gathered}$$

Multiply $\left(iii\right)$ by $2$, We get,

$4x+6y=8...\left(iii\right)$

Step 2. subtract $\left(iv\right)$ from $\left(iii\right)$ and solving, we get

$4x+6y-8-4x-5y+6=0$

$\Rightarrow$ $0x+1y-2=0$

$\Rightarrow$ $0x+1y=2$

$\Rightarrow$ $0+y=2$

$\Rightarrow$ $y=2$

put $\text{'}y\text{'}$ value in equation $\left(i\right)$

$\Rightarrow$ $x+2\left(2\right)=3$

$\Rightarrow$ $x+4=3$

$\Rightarrow$ $x=3-4$

$\Rightarrow$ $x=-1$

$\Rightarrow$ $x=-1,$$y=2.$

Step 3. Put $\left(-1,2\right)$ in equation $\left(iii\right)$

$\Rightarrow$$4\left(-1\right)+6\left(2\right)=8$

$\Rightarrow$ $-4+12=8$

$\Rightarrow$ $8=8$

$\left(-1,2\right)$ satisfies (iii)

Step 4. Put $\left(-1,2\right)$ in equation$\left(iv\right)$

$\Rightarrow$$4\left(-1\right)+5\left(2\right)=6$

$\Rightarrow$ $-4+10=6$

$\Rightarrow$ $6=6$

$\left(-1,2\right)$ satisfies$\left(iv\right)$

$\therefore$$\left(iii\right)$and $\left(iv\right)$ intersect at $(-1,2)$

Step 5. put $\left(-1,2\right)$ in equation$\left(i\right)$

$\Rightarrow$$-1+2\left(2\right)=3$

$\Rightarrow$ $-1+4=3$

$\Rightarrow$ $3=3$

$\left(-1,2\right)$ satisfies $\left(i\right)$

So this line will also pass through the point $\left(-1,2\right)$

Step 6. put $\left(-1,2\right)$ in equation$\left(ii\right)$

$\Rightarrow$$3(-1)+4\left(2\right)=7$

$\Rightarrow$ $-3+8=7$

$\Rightarrow$ $5\ne 7$

$\left(-1,2\right)$ not satisfies$\left(ii\right)$.

so this line will not pass through the point $\left(-1,2\right).$

$\therefore$Only three lines are concurrent.

Hence, option(D) is correct.