Question

If the point \([1+(t / \sqrt{2}), 2+(t / \sqrt{2})]\) lies between the two parallel lines \(x+2 y=1\) and \(2 x+4 y=15\), then the range of \(t\) is \(\ldots \ldots\) (a) \(0

Short answer

The range of \(t\) is \(-\frac{4\sqrt{2}}{3}<t<\frac{5\sqrt{2}}{6}\).

Step-by-step solution

Substitute the point into the equations of the lines

First, we need to substitute the coordinates of the point \([1+(\frac{t}{\sqrt{2}}), 2+(\frac{t}{\sqrt{2}})]\) into the equations of both lines. For the first line, \(x + 2y = 1\), \[1 + (\frac{t}{\sqrt{2}}) + 2[2 + (\frac{t}{\sqrt{2}})] = 1\] For the second line, \(2x + 4y = 15\), \[2[1 + (\frac{t}{\sqrt{2}})] + 4[2 + (\frac{t}{\sqrt{2}})] = 15\]

Simplify the equations

Now, we will simplify both equations to find the relationship between the \(t\) and the lines. For the first equation, \[1 + (\frac{t}{\sqrt{2}}) + 4 + (\frac{2t}{\sqrt{2}}) = 1\] \[\frac{3t}{\sqrt{2}} = -4\] For the second equation, \[2 + (\frac{2t}{\sqrt{2}}) + 8 + (\frac{4t}{\sqrt{2}}) = 15\] \[\frac{6t}{\sqrt{2}} = 5\]

Solve for t

We will solve the two equations by isolating \(t\). From the first equation, \[t = -\frac{4\sqrt{2}}{3}\] From the second equation, \[t = \frac{5\sqrt{2}}{6}\]

Find the range of t

Since the point lies between the two lines, we need \(-\frac{4\sqrt{2}}{3} < t < \frac{5\sqrt{2}}{6}\). And this matches the answer choice (c), so the range of \(t\) is \(\boxed{-\frac{4\sqrt{2}}{3}<t<\frac{5\sqrt{2}}{6}}\).

Key concepts

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is a powerful tool for analyzing geometric properties and relationships by means of algebraic equations. In essence, it bridges the gap between algebra and geometry, allowing complex problems, such as finding the range of variables constrained by geometric context, to be solved more easily.

To get a better grasp of coordinate geometry, let's start by understanding the Cartesian coordinate system. It consists of two perpendicular axes: the horizontal axis (x-axis) and the vertical axis (y-axis). The point where these axes intersect is known as the origin. Any location in the plane can be identified by an ordered pair \( (x, y) \) of numbers, which represent its distances from the two axes.

Working with Points and Lines

In coordinate geometry, the position of a point is based on its x and y coordinates, while a line’s properties are described by an equation. For instance, if you have a point \( (x_{1}, y_{1}) \) and you want to test whether it lies between two lines, you need to substitute its coordinates into the equations of the lines. If the point satisfies both equations, it may indicate that it lies on one line or the other, or within the region those lines bound.In the exercise provided, a point with coordinates dependent on \( t \) is given, and we are asked to find the range of \( t \) so that the point lies between two parallel lines. This requires an understanding of how to substitute these coordinates into the lines' equations and how to interpret the inequalities that arise.

Equation of a Line

An equation of a line provides a recipe for finding all the points that lie on that line. In two dimensions, the simplest form of a line's equation is \( y = mx + b \) where \( m \) represents the slope of the line, and \( b \) represents the y-intercept, which is the point where the line crosses the y-axis.

There are other forms of a line's equation as well, such as the standard form \( Ax + By = C \) and the point-slope form \( y - y_{1} = m(x - x_{1}) \) where \( (x_{1}, y_{1}) \) is a known point on the line. Understanding these various forms and how to manipulate them to solve for unknowns within them is a cornerstone of coordinate geometry.

Analyzing Linear Equations

In our textbook exercise, we have two equations of parallel lines. Lines are parallel when they have the same slope, which in standard form can be seen by the ratio of the coefficients of \( x \) and \( y \) being equal. We need to interpret these equations together with the point given to find the correct range of \( t \). This involves rearranging and simplifying the equations as demonstrated in the solution steps, and then using that information to draw a conclusion about the value of \( t \) that fits within the constraints of the problem.

Linear Inequalities

Linear inequalities are similar to equations of a line, but instead of equalities, they express ranges of values where one expression is greater than or less than another. These can describe regions of the coordinate plane where a particular condition is met, restricting a variable's possible values.

An inequality such as \( ax + by < c \) depicts a half-plane, where all points that satisfy this condition lie on one side of the boundary line defined by \( ax + by = c \) and not on the line itself. To solve linear inequalities, one often performs the same operations as with linear equations, but with careful attention to how these operations affect the inequality's direction, especially multiplication or division by a negative number.

Applying Inequalities to the Range of Variables

The end goal in our original problem is to find the range of the variable \( t \) so that the point lies within the bounds set by the parallel lines. After substituting the point's coordinates into the line equations and finding equivalent expressions involving \( t \) only, we end up with two inequalities, which, when combined, provide the solution to the problem. Understanding the relationship between inequalities and the geometrical representation of lines empowers students to visualize and solve for variables constrained by geometric conditions.