 Hi, let us discuss the following question. The question says, reduce the following equations into normal form, find the perpendicular distances from the origin and angle between perpendicular and the positive x-axis. Second part is, y minus 2 is equal to 0. Before solving this question, we should know that normal form of equation Ax plus By plus C is equal to 0 is given by x cos omega plus y sine omega is equal to p, where p is the perpendicular distance from the origin and omega is the angle between perpendicular and the positive x-axis. Let's now begin with the solution. Given equation S, y minus 2 is equal to 0. Now this equation can be written as 0 into x plus 1 into y minus 2 is equal to 0. Now this implies 0 into x plus 1 into y is equal to 2. We have shifted the constant to the right-hand side. Now we will divide both sides of this equation by square root of coefficient of x, which is 0 here, plus coefficient of y whole square, and here coefficient of y is 1. So this is equal to 1. So on dividing both sides of this equation by 1, we get 0 by 1 into x plus 1 into y is equal to 2. Now cos 90 degree is equal to 0, and sine 90 degree is equal to 1. So this equation can be written as x cos 90 degree plus y sine 90 degree is equal to 2. So after reducing the given equation into normal form, we have got x cos 90 degree plus y sine 90 degree is equal to 2. Here the perpendicular distance from the origin, that is, p is equal to 2. And the angle between perpendicular and the positive x-axis, that is, omega is equal to 90 degree. Hence the required normal form of the given equation is x cos 90 degree plus y sine 90 degree is equal to 2. Propenticular distance from the origin is 2. And the angle between perpendicular and x-axis is 90 degree. This is our required answer. So this completes the session. Bye and take care.