If the point \([1+(t / \sqrt{2}), 2+(t / \sqrt{2})]\) lies between the two parallel lines \(\mathrm{x}+2 \mathrm{y}=1\) and \(2 \mathrm{x}+4 \mathrm{y}=15\), then the range of \(\mathrm{t}\) is \(\ldots \ldots\) (a) \(0<\mathrm{t}<[5 /(6 \sqrt{2})]\) (b) \(-[(4 \sqrt{2}) / 3]<\mathrm{t}<0\) (c) \(-[(4 \sqrt{2}) / 3]<\mathrm{t}<[(5 \sqrt{2}) / 6]\) (d) None of these

Coordinate geometry, also known as analytic geometry, is a system of geometry where the position of points on the plane is described using an ordered pair of numbers, typically known as coordinates. The basic element of coordinate geometry is the concept of graphing lines and curves through equations.

When it comes to the equation of a line in the plane, it can be represented in various forms such as slope-intercept form or standard form. In our exercise, the lines are given in the form of linear equations, and the coordinates of a point are given, which we are trying to place with respect to these lines. Using the principles of coordinate geometry, we can determine if the point lies between the lines, on the lines, or outside of them by simply plugging the point's coordinates into the equations of the lines and analysing the resulting inequalities.

Linear equations represent straight lines in coordinate geometry and are an essential concept in mathematics. The standard form of a linear equation is given as \( Ax + By = C \) where \( A \) and \( B \) are not both zero. The solution to a linear equation is the set of all points \( (x, y) \) that make the equation true.

In our exercise, we have utilized two linear equations, which represent two distinct parallel lines. Parallel lines never intersect and have the same slope. The coefficients of \( x \) and \( y \) in both equations reveal this similarity in slope. By substituting the coordinates of the given point into these equations, one can solve for the variable \( t \) to find the values that satisfy the equations, ultimately determining the position of the point and the range of \( t \) for which the point will lie between these two parallel lines.

While the exercise does not directly involve the solution of triangles, the methods of analytical geometry apply to various geometrical configurations. The solution of triangles refers to finding unknown sides and angles in a triangle when certain sides and angles are known using trigonometric methods.

However, we can draw parallels between the concept and our exercise. For instance, the point that lies between two parallel lines splits the strip of the plane between these lines similarly to how a median might split a triangle into smaller sections. Such methodologies, along with the understanding of how linear equations work, help us better visualize geometric scenarios and lead us to relevant solutions.

JEE Maths is a critical part of the Joint Entrance Examination (JEE), which is a standardized test taken by engineering aspirants in India. It includes a wide range of topics from higher secondary mathematics, such as coordinate geometry and linear equations, which we encounter in this problem.

For students preparing for the JEE, mastering the ability to translate a geometrical problem into algebraic form and vice versa is quintessential. This exercise is a prime example of such a problem where understanding the concept is not just about solving the equations but also about interpreting the geometric situation presented. The question's format and the approach to finding the range of \( t \) reflect the kind of analytical thinking expected in JEE Maths. Engaging with these problems enhances problem-solving skills and prepares a student for the range of questions that the JEE might present.