Question:

## **Introduction** In **linear algebra and coordinate geometry**, a **reflection** is a linear transformation that maps every point in the plane to its mirror image across a given line. Such transformations can be represented compactly using a **transformation matrix**, which allows us to reflect any point (or geometric figure such as a triangle) by simple matrix multiplication. In this problem, a triangle with known vertices is reflected about a straight line, and we are required to **find the transformation matrix** corresponding to this reflection. --- ## **Given Data** Triangle (ABC) has vertices: [ A(3,3), \quad B(3,5), \quad C(5,3) ] The line of reflection is: [ y = -\frac{x}{10} ] This line passes through the origin and has slope: [ m = -\frac{1}{10} ] --- ## **Theory: Reflection about a Line Through the Origin** For a line through the origin making an angle ( \theta ) with the positive (x)-axis, the **reflection matrix** is: [ R = \begin{bmatrix} \cos 2\theta & \sin 2\theta \ \sin 2\theta & -\cos 2\theta \end{bmatrix} ] The slope of the line is related to ( \theta ) by: [ \tan \theta = m ] --- ## **Step 1: Determine ( \theta )** [ \tan \theta = -\frac{1}{10} ] Using trigonometric identities: [ \cos 2\theta = \frac{1 - \tan^2\theta}{1 + \tan^2\theta}, \quad \sin 2\theta = \frac{2\tan\theta}{1 + \tan^2\theta} ] Substitute ( \tan\theta = -\frac{1}{10} ): [ \tan^2\theta = \frac{1}{100} ] [ \cos 2\theta = \frac{1 - \frac{1}{100}}{1 + \frac{1}{100}} = \frac{\frac{99}{100}}{\frac{101}{100}} = \frac{99}{101} ] [ \sin 2\theta = \frac{2(-\frac{1}{10})}{1 + \frac{1}{100}} = \frac{-\frac{1}{5}}{\frac{101}{100}} = -\frac{20}{101} ] --- ## **Step 2: Construct the Reflection Matrix** [ R = \begin{bmatrix} \frac{99}{101} & -\frac{20}{101} \ -\frac{20}{101} & -\frac{99}{101} \end{bmatrix} ] --- ## **Final Answer** The **transformation matrix** that reflects any point (and hence triangle (ABC)) about the line [ y = -\frac{x}{10} ] is: [ \boxed{ \begin{bmatrix} \frac{99}{101} & -\frac{20}{101} \ -\frac{20}{101} & -\frac{99}{101} \end{bmatrix} } ] --- ## **Conclusion** Reflection about an arbitrary line through the origin can be represented using a trigonometric-based transformation matrix. By relating the slope of the line to the angle it makes with the (x)-axis, we derived the required matrix. This matrix can now be applied to each vertex of triangle (ABC) to obtain its reflected image.