We read your photo as:

Find the equation of the line of reflection for the point (3,3) reflected to (3,-3).

Mathematics

ScanSolve AI Answer

Answer

0x+y+0=00x+y+0=0

Step-by-step solution

5 steps

  1. Find the midpoint of the segment AA'.

    The coordinates of point A are (3,3)(3,3) and its image A' are (3,−3)(3,-3).
    The midpoint M of the segment AA' is given by the formula:

    M=(x1+x22,y1+y22)M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)

    Substitute the coordinates of A and A':

    M=(3+32,3+(−3)2)M = \left(\frac{3+3}{2}, \frac{3+(-3)}{2}\right) M=(62,02)M = \left(\frac{6}{2}, \frac{0}{2}\right) M=(3,0)M = (3,0)

  2. Find the slope of the segment AA'.

    The slope mAA′m_{AA'} of the segment connecting A(3,3)(3,3) and A'(3,−3)(3,-3) is given by the formula:

    mAA′=y2−y1x2−x1m_{AA'} = \frac{y_2-y_1}{x_2-x_1} mAA′=−3−33−3m_{AA'} = \frac{-3-3}{3-3} mAA′=−60m_{AA'} = \frac{-6}{0}

    The slope is undefined, which means the segment AA' is a vertical line.

  3. Determine the slope of the mirror line.

    The mirror line is the perpendicular bisector of the segment AA'. Since AA' is a vertical line, the mirror line must be a horizontal line.
    The slope of a horizontal line is 0.

  4. Find the equation of the mirror line.

    The mirror line passes through the midpoint M(3,0)(3,0) and has a slope of 0.
    Using the point-slope form of a linear equation, y−y1=m(x−x1)y - y_1 = m(x - x_1):

    y−0=0(x−3)y - 0 = 0(x - 3) y=0y = 0

  5. Express the equation in the form ax+by+c=0ax+by+c=0.

    The equation y=0y=0 can be written as:

    0x+1y+0=00x + 1y + 0 = 0

    So, the equation of the mirror line is 0x+y+0=00x+y+0=0.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Mathematics Questions

QFind the equation of a line parallel to 2y = 3(x − 2) and passing through the point (2, 3).QFind the equation of the tangent and normal lines to the curve y = x3 - 3x − 1 at the point (1, −3).QFind the slope of the line that passes through the points (2, −3) and (3, 4).QThe rule for reflection across the x-axis is (x, y) (x, −y).QFind the equation of a straight line whose gradient is 3 and passes through point C(−2, 5).
See all Mathematics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp