 Welcome to the session. Let us discuss the following question. The question says find the distance of the point minus 1, 1 from the line 12 into x plus 6 is equal to 5 into y minus 2. Before solving this question, we should know the formula for finding the distance of a point from the line. v, y plus c is equal to 0 is the equation of the line and p is the point having coordinates x1, y1. The distance of this point from this line will be given by mod of a into x1 plus v into y1 plus c upon square root of a square plus v square. So always remember the distance of a point from a line is given by mod of a x1 plus v by 1 plus c upon square root of a square plus v square. Next now begin with the solution. Given equation of line as 12 into x plus 6 is equal to 5 into y minus 2. Now this implies 12x plus 72 is equal to 5, 5 minus 10. This implies 12x minus 5, 5 plus 72 plus 10 is equal to 0. This implies 12x minus 5, 5 plus 82 is equal to 0. Now this equation is of the form a x plus v y plus c is equal to 0. On comparing this equation with a x plus v y plus c is equal to 0, we find that a is equal to 12, b is equal to minus 5 and c is equal to 82. We have to find the distance of the point minus 1, 1 from this line that is 12x minus 5, 5 plus 82 is equal to 0. Now here the point is minus 1, 1. So this means x1, y1 is equal to minus 1, 1. We have learnt above that distance of a point from the line is given by mod of a x1 plus b y1 plus c upon square root of a square plus b square. Right? So now we will substitute the value of a, b, c and x1, y1 in this formula. By substituting the values we get the required distance as mod of 12 into minus 1 plus minus 5 into 1 plus 82 upon square root of 12 square plus minus 5 square. Now this is equal to minus 12 minus 5 plus 82 upon 1.4 plus 25 and this is equal to mod of 65 upon 169 and this is equal to 65 by 30 and this is equal to 5. Hence the required distance is 5 units. This is our required answer so this completes the session by J. Gill.