Solution :
Solving simultaneously the equations 2x – y + 3 = 0 and x + y – 5 = 0, we obtain
\(x\over {5 – 3}\) = \(y\over {3 + 10}\) = \(1\over {2 + 1}\)
\(\implies\) \(x\over 2\) = \(y\over 13\) = \(1\over 3\)
\(\implies\) x = \(2\over 3\) , y = \(13\over 3\)
Hence, (2/3, 13/3) is the required point of intersection.
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The slope of tangent parallel to the chord joining the points (2, -3) and (3, 4) is