Find a point symmetrical to the point p (1, -1) relative to the line 3x-5y + 9 = 0.

We have a straight line:

3 * x – 5 * y + 9 = 0;

Let us bring it to the form y = k * x + b:

5 * y = 3 * x + 9;

y = 3/5 * x + 9/5;

A straight line perpendicular to the given one will have the opposite slope:

k2 = -k1 = -3/5;

y = -3/5 * x + b;

Substitute the point values into the formula:

-1 = -3/5 + b;

b = -2/5;

y = -3/5 * x – 2/5 – equation of a straight line.

Intersection point of lines:

3/5 * x + 9/5 = -3/5 * x – 2/5;

6/5 * x = -11/5;

x = -11/6;

A point symmetrical to a given point has an abscissa:

x0 = -11/6 – (1 + 11/6) = -11/6 – 17/6 = -28/6 = -14/3;

y0 = -3/5 * (-14/3) – 2/5 = 14/5 – 2/5 = 12/5;