# Image of a Point in a Plane – Formula and Example

Here you will learn how to find image of a point in a plane formula with examples.

Let’s begin –

## Image of a Point in a Plane

Let P and Q be two points and let $$\pi$$ be a plane such that

(i)  line PQ is perpendicular to the plane $$\pi$$ and,

(ii) mid-point of PQ lies on the plane $$\pi$$.

Then, either of the point is the image of the other in the plane $$\pi$$.

In order to find the image of a points $$(x_1, y_1, z_1)$$ in the plane ax + by + cz + d = 0, we may use the following algorithm.

Algorithm :

1). Write the equations of the line passing through P and normal to the given plane as $$x – x_1\over a$$ = $$y – y_1\over b$$ = $$z – z_1\over c$$.

2). Write the coordinates of image Q as $$(x_1 + ar, y_1 + br, z_1 + cr)$$.

3). Find the coordinates of the mid-point R of PQ.

4). Obtain the value of r by substituting the coordinates of R in the equation of the plane.

5). Put the value of r in the coordinates of Q.

Example : Find the image of the point P (3, -2, 1) in the plane 3x – y + 4z = 2.

Solution : Let Q be the image of P(3, -2, 1) in the plane 3x – y + 4z = 2. Then PQ is normal to the plane. Therefore, direction ratios of PQ are proportional to 3, -1, 4. Since PQ passes through P(3, -2, 1) and has direction ratios proportional to 3, -1, 4.

Therefore, equation of PQ is

$$x – 3\over 3$$ = $$y + 2\over -1$$ = $$z – 1\over 4$$ = r   (say)

Let the coordinates of Q be (3r + 3, -r – 2, 4r + 1). Let R be the mid-point of PQ. Then, R lies on the plane 3x – y + 4z = 2.

The coordinates of R are ($${3r + 3 + 3\over 3}, {-r – 2 – 2\over 2}, {4r + 1 +1\over 2}$$)

or,  ($${3r + 6\over 3}, {-r – 4\over 2}, {2r + 1}$$)

Since r lies on the plane 3x – y + 4z = 2.

$$3({3r + 6\over 2}) – ({-r – 4\over 2})$$ + 4(2r + 1) = 2

$$\implies$$ 13r = -13 $$\implies$$ r = – 1

Putting r = -1 in (3r + 3, -r – 2, 4r + 1), we obtain the coordinates of Q as (0, – 1, -3).

Hence, the image of (3, -2, 1) in the plane 3x –  y + 4z = 2 is (0, -1, -3).