Region R is a square in the x-y plane with vertices

The square is formed by x=5 & x=-1 ; y=4 & y=-2

-1 to 5 or -2 to 4 = 6 which is the length of the square (if it was an rectangle we would have to find the length and breadth)

So the area of the square is 36.

To find the points where the line 3x-5y=10 cuts across the square, insert the two values of x or y into the equation

x=5
=> 3*5-5y=10
=> y=1. So (5,1) is first point of intersection of the line with the square.

x=-1
=> 3*(-1)-5y=10
=> y=-(7/5). This point (-1, -7/5) is not on any of the boundary of the square so this is not relevant. Let's try inserting the values of y to find where is the second point of intersection is

y=-2
=> 3x-5*(-2)=10
=> x=0. So (0, -2) is the second point of intersection of the line with the square.

So the line cuts the square at two adjacent sides of the square to form a triangle with base 5 and height 3. Area of this triangle is 15/2

Probability of point lying in this triangle = (15/2)/36 = 5/24