In the line \(AB\) by visualization point \(A\) is \((x,0)\).

and point \(B\) is \((0,y)\)

Since the given point \(c(-3,5)\) is the mid-point of the line we can make a eqn

\(x + 0/2 = -3\)

or, \(x = -6\)

\(\frac{y + 0}{2} = 5\)

therefore \(y = 10\)

For line \(AB\) \(A(-6,0) B(0,10)\)

Now, we can similarly calculate the value of point \(D\) and \(E\) in the line \(DE\)

Point \(D(0,y)\) and point \(E(x,0)\)

Hence, \(\frac{0+x}{2} = 3 or x = 6\)

\(\frac{y+0}{2} = -5\)

or, \(y=-10\)

For line \(DE\) D(0,-10) and \(E(6,0)\)

when we have two points we can calculate the slope for each line AB and DE we have three set of points so we can calculate slope

For the line \(AB\) slope \(= \frac{10-0}{0-(-6)} = \frac{10}{6} = \frac{5}{3}\)

For the line \(DE\) slope \(= \frac{0-(-10)}{6-0} = \frac{5}{3}\)

option c

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