Given four lines with equations $x + 2 y - 3 = 0, 3 x + 4 y - 7 = 0$ , $2 x + 3 y - 4 = 0, 4 x + 5 y - 6 = 0$ , then they are all concurrent

True

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False

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Solution

The correct option is B False
The point of intersection of
$x + 2 y - 3 = 0$ and $2 x + 3 y - 4 = 0$
is found to be $(- 1, 2)$ . The co-ordinates of this point satisfy the equation $4 x + 5 y - 6 = 0$ but not $3 x + 4 y - 7 = 0$ . Hence all the four lines are not 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

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

2 x + 3 y - 4 = 0

3 x + 4 y - 5 = 0

4 x + 5 y - 6 = 0

x^{2} + y^{2} - 6 x + 4 y = 12

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