# Find the image of the point (1,2) in the line x-2y-7= 0 .

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You need to find the reflection of the given point (1,2), over the line `x - 2y - 7 = 0.`

You need to drop a perpendicular from the point `(1,2), ` to the given line. You should come up with the notation M for the intersection point between the perpendicular line and the given line.

M represents the midpoint of the segment whose endpoints are (1,2) and its reflection.

You need to write the equation of the perpendicular line, passing through the point (1,2), using the following equation:

`y - 2 = m_2(x -1)`

You may find the slope m using the relation between the slopes of two perpendicular lines such that:

`m_1*m_2 = -1`

You need to convert the general form of the given equation of line into slope intercept form such that:

`-2y = -x - 7 => 2y = x + 7 => y = x/2 + 7/2 => m_1 = 1/2`

`m_2 = -2`

Substituting -2 for `m_2` in equation of perpendicular line yields:

`y - 2 = -2(x -1) => y = -2x + 4`

You should find the point of intersection M solving the system of equations such that:

`{(y = x/2 + 7/2),(y = -2x + 4):} =>x/2 + 7/2 = -2x + 4 => 2x + x/2 = 4 - 7/2 => 5x = 1 => x =1/5`

`y = -2/5 + 4 => y = 18/5`

Hence, evaluating the coordinates of midpoint M yields: M(`1/5;18/5` )

You need to use the formula of midpoint to evaluate the image of the point (1,2) such that:

`x_M = (1 + x)/2 => 2/5 = 1 + x => 2/5 - 1 = x => x = -3/5`

`y_M = (1 + y)/2 =>36/5 =1 + y => y = 31/5`

**Hence, evaluating the reflection of the given point over the given line yields (-`3/5 ; 31/5` ).**

**Sources:**

Its Answer Is (5,-6). This question is from Book Understanding ICSE Mathematics by M.L.Aggarwal and this question is from chapter Equation of straight line .