Referencing the diagram shown below, where the origin is at the lower left hand corner of the rectangle:
Let's say the coordinates of the top point are $(a,b)$ and that $q=(x_q,y_q)$ is the point of intersection "on the left" of the diagonal and the rectangle.
Draw the right triangle whose hypotenuse is the line segment between the points $(a,b)$ and $(240,150)$, whose vertical side intersects $(a,b)$, and whose horizontal side intersects $(240,150)$. Let's call that triangle $T$.
Draw the right triangle whose hypotenuse is the line segment between the points $(a,b)$ and $q$, whose vertical side intersects $(240,150)$, and whose horizontal side intersects $q$. Let's call that triangle $B$.
For triangle $T$, we know the height of the vertical side is $|b-150|$. We also know the width of the horizontal side $|a-240|$.
For triangle $B$, we know the height of the vertical side is $ |150 -q_y|$. And we know the width of the horizontal side $ |240-q_x|$.
These two triangles are similar, so corresponding ratios of sides are equal: $ \tag{1}{|b-150|\over |a-240|}={ |150-q_y|\over|240-q_x|}. $
Now, depending on whether the diagonal intersects the top/bottom or left/right sides of the rectangle, you will either know the value of $q_x$ or $q_y$. Then you can use equation (1) to find the value of the other coordinate.
Of course, you could just find the equation of the line formed by the diagonal and use that...
In fact, the equation would be $ \tag{2}{y-150\over x-240}={b-150\over a-240}. $
You know either the $x$ or $y$ coordinate of $q$; so, you can use equation (2) to find the unknown coordinate.
(Note, incidentally, the similar triangle argument essentially tells you why the equation of the line is as given in (2).)
Not to scale