The four lines with equations $x + 2 y - 3 = 0, 3 x + 4 y - 7 = 0, 2 x + 3 y - 4 = 0, 4 x + 5 y - 6 = 0$

are all concurrent

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have three lines which are concurrent

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do not enclose a quadrilateral

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have two lines that pass through

(1, 1)

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Solution

The correct options are
A have three lines which are concurrent
C do not enclose a quadrilateral
D have two lines that pass through $(1, 1)$
Given lines, $x + 2 y - 3 = 0$ and $3 x + 4 y - 7 = 0$ intersect at $(1, 1),$ which does not satisfy $2 x + 3 y - 4 = 0$ and $4 x + 5 y - 6 = 0$ .
Also, $3 x + 4 y - 7 = 0$ and $2 x + 3 y - 4 = 0$ intersect at $(5, - 2)$ which does not satisfy $x + 2 y - 3 = 0$ and $4 x + 5 y - 6 = 0.$
Lastly, intersection point of $x + 2 y - 3 = 0$ and $2 x + 3 y - 4 = 0$ is $(- 1, 2)$ which satisfy $4 x + 5 y - 6 = 0.$
Hence, only there lines are concurrent.

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Similar questions

x + 2 y - 3 = 0, 3 x + 4 y - 7 = 0

2 x + 3 y - 4 = 0, 4 x + 5 y - 6 = 0

\begin{matrix} x + 2 y - 3 = 0, 3 x + 4 y - 7 = 0 2 x + 3 y - 4 = 0, 4 x + 5 y - 6 = 0 \end{matrix}

x + 2 y - 3 = 0, 3 x + 4 y - 7 = 0

2 x + 3 y - 4 = 0, 4 x + 5 y - 6 = 0

x + 2 y - 3 = 0

2 x + 3 y - 4 = 0

3 x + 4 y - 5 = 0

4 x + 5 y - 6 = 0

x - y - 5 = 0

{(\frac{15}{A B})}^{2} + {(\frac{10}{A C})}^{2} = {(\frac{6}{A D})}^{2}

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