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  4. The lines x + 2y - 5 = 0, 2x - 3y + 4 = 0, 6x + 4y - 13 = 0

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Q. The lines x + 2y - 5 = 0, 2x - 3y + 4 = 0, 6x + 4y - 13 = 0

Straight Lines

Solution:

Lines II and III are at right angles
$[ \because \, (\frac{2}{3})(- \frac{3}{2}) = - 1$
Lines I and II intersect at the point (1, 2) and (1, 2) does not belong to III. Hence, the lines are not concurrent, i.e., they form a right angled triangle.