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xleb123 · Accepted Answer

To find the images of the points P(3,2), Q(0,1), and R(-5,7) when reflected about the line

y = \frac{1}{\sqrt{3}} x

, we can use the formula for the reflection of a point across a line.
Let's denote the line as

L : y = \frac{1}{\sqrt{3}} x

.
The formula for reflecting a point (x, y) across a line can be expressed as follows:
1. Find the slope of the line perpendicular to L. Let's call this slope m. The slope of a line perpendicular to L is the negative reciprocal of the slope of L.
2. Find the equation of the perpendicular line passing through the given point.
3. Find the intersection point of L and the perpendicular line. This point is equidistant from the given point and its reflection.
4. Calculate the distance between the given point and the intersection point.
5. Finally, find the reflection point by using the distance and the intersection point.
Now let's apply this procedure to each point:
a) Point P(3, 2):
The slope of the line L is

\frac{1}{\sqrt{3}}

. Therefore, the slope of the perpendicular line is

- \sqrt{3}

.
Using the point-slope form, we can find the equation of the perpendicular line passing through P(3, 2):

y - 2 = - \sqrt{3} (x - 3)

Simplifying, we get:

y = - \sqrt{3} x + 9 + 2

y = - \sqrt{3} x + 11

To find the intersection point of L and the perpendicular line, we solve the system of equations:

{\begin{matrix} y = \frac{1}{\sqrt{3}} x \\ y = - \sqrt{3} x + 11 \end{matrix}

Solving this system, we find the intersection point to be

(\frac{3}{4}, \frac{3}{4})

.
Now, let's calculate the distance between P(3, 2) and the intersection point:

d = \sqrt{{(\frac{3}{4} - 3)}^{2} + {(\frac{3}{4} - 2)}^{2}} = \sqrt{\frac{45}{16}}

The reflection point, P', is located on the other side of the line L and is equidistant from P and the intersection point. Therefore, the coordinates of P' are:

(\frac{3}{4} - \sqrt{\frac{45}{16}}, \frac{3}{4} - \sqrt{\frac{45}{16}})

b) Point Q(0, 1):
Following the same steps as before, we find that the reflection point Q' is located at

(- \frac{1}{4} - \sqrt{\frac{15}{16}}, - \frac{1}{4} + \sqrt{\frac{15}{16}})

.
c) Point R(-5, 7):
Similarly, the reflection point R' is located at

(- \frac{1}{4} + \sqrt{\frac{135}{16}}, - \frac{1}{4} - \sqrt{\frac{135}{16}})

.
Therefore, the images of the points P(3, 2), Q(0, 1), and R(-5, 7) when reflected about the line

y = \frac{1}{\sqrt{3}} x

are:
P'

(\frac{3}{4} - \sqrt{\frac{45}{16}}, \frac{3}{4} - \sqrt{\frac{45}{16}})

Q'

(- \frac{1}{4} - \sqrt{\frac{15}{16}}, - \frac{1}{4} + \sqrt{\frac{15}{16}})

R'

(- \frac{1}{4} + \sqrt{\frac{135}{16}}, - \frac{1}{4} - \sqrt{\frac{135}{16}})

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