 Hello and welcome to the session. In this session we will discuss the question which says that reducing this straight line phi x plus 4 y plus 5 phi is equal to 0 to its normal form, find the perpendicular distance of the line from the region. Now before starting the solution of this question, we should know a result and that is equation of the straight line in normal form. Now the equation of a straight line in normal form is cos alpha plus y sin alpha is equal to p where is the length of the perpendicular from the region to the line. Also it is always measured away from the region and it is always positive in value. And alpha is a positive angle less than 360 degrees is measured from the positive direction of x axis to the normal from the region to the line. So this result will work out as a key idea for solving out this question. And now we will start with the solution. Here the given equation is 3x plus 4y plus 15 is equal to 0. This implies minus 3x minus 4y is equal to 15. Now let us name it as 1. Now we know that the equation of the straight line in normal form is x cos alpha plus y sin alpha is equal to p. So we have x cos alpha plus y sin alpha is equal to p. Now comparing one with this equation we get cos alpha over minus 3 is equal to sin alpha over minus 4 is equal to p over 15. Now from these two we get cos alpha over minus 3 is equal to p over 15 which implies cos alpha is equal to minus 3p by 15. Also from these two we have sin alpha over minus 4 is equal to p by 15. This implies is equal to minus 4p by 15. Now we know that for trigonometric identity cos square alpha plus sin square alpha is equal to 1. Now putting the values of cos alpha and sin alpha here this implies 3p by 15 whole square plus minus 4p by 15 whole square is equal to 1. This implies 9p square over 225 plus 16p square 25 is equal to 1. Further taking the same as 225 this will be 9p square p square is equal to 1. This whole square 16p square is 25p square over 225 is equal to 1. This implies 1 into 9 is 225 so it will be is equal to 1. 9 which implies we are only taking the positive value as we know that is always positive is equal to minus 3p by 15. So putting the value of p here it will be minus 3 into 3 over 15 which is equal to minus 3 by 5. Also is equal to minus 4p by 15 so putting the value of p here it will be minus 4 into 3 by 15 which is equal to minus 4 by 5. Now we know that this trigline in a normal form is x square alpha plus y sin alpha is equal to p. So putting the value of cos alpha, sin alpha and p here this implies minus 3 by 5. Further this implies alpha by 5 is equal to 3 on both side this implies 4y is equal to minus 15. Further this implies 3x plus 4y plus 15 is equal to 0 of the trigline in the normal form. And the perpendicular distance is given by p and here p is equal to 3 solution of the given question. And that's all for this session hope you all have enjoyed the session.