Coordinates of Points on Line at Fixed Distance

This problem asks us to find the coordinates of two points, P and Q. These points must satisfy two conditions simultaneously: they must lie on the straight line defined by the equation $y = 2x + 3$, and they must be exactly $2$ units away from the given point $R(1, 5)$.

Problem Conditions Analysis

• Condition 1: Points lie on the line
  Any point $(x, y)$ that lies on the line $y = 2x + 3$ must satisfy this equation. This means if we know the x-coordinate, we can find the y-coordinate, and vice versa.
• Condition 2: Fixed distance from a point
  The points P and Q are at a distance of $2$ units from the point $R(1, 5)$. The distance formula is used to calculate the distance between two points in a coordinate plane.

Equations Setup

Let $P(x, y)$ be one of the points we are looking for.

From Condition 1, the coordinates $(x, y)$ must satisfy:

From Condition 2, the distance between $P(x, y)$ and $R(1, 5)$ is $2$. Using the distance formula, $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, we have:

To simplify calculations, we can square both sides of the distance equation:

Solving Coordinates

Now, we need to use both equations together. We substitute the expression for $y$ from the line equation ($y = 2x + 3$) into the squared distance equation:

Simplify the term inside the second parenthesis:

Notice that $(2x - 2)$ can be written as $2(x - 1)$. Substitute this back into the equation:

Expand the second term:

Combine the like terms:

Now, isolate $\left(x - 1\right)^2$:

Take the square root of both sides to solve for $(x - 1)$:

Solve for $x$. This gives us two possible values for the x-coordinate:

Y-values Calculation

Now we find the corresponding y-coordinates for each x-value using the line equation $y = 2x + 3$.

Point P Calculation

For the first x-value, $x_1 = 1 + \frac{2}{\sqrt{5}}$:

So, one point is $\left(1 + \frac{2}{\sqrt{5}}, 5 + \frac{4}{\sqrt{5}}\right)$.

Point Q Calculation

For the second x-value, $x_2 = 1 - \frac{2}{\sqrt{5}}$:

So, the other point is $\left(1 - \frac{2}{\sqrt{5}}, 5 - \frac{4}{\sqrt{5}}\right)$.

Points P and Q Coordinates

The coordinates of the two points P and Q that lie on the line $y = 2x + 3$ and are $2$ units away from the point $R(1, 5)$ are:

Point P: $(1+\frac{2}{\sqrt{5}},5+\frac{4}{\sqrt{5}})\$$

Point Q: $(1-\frac{2}{\sqrt{5}},5-\frac{4}{\sqrt{5}})\$$