 Hi, and welcome to the session. Let us discuss the following question. The question says find the image of the point 3A with respect to the line x plus 3y is equal to 7, assuming the line to be a plane error. Let's make a figure to understand this question. Let AB be the line mirror whose equation is x plus 3y is equal to 7. Let P be a point having coordinates 3A. We have to find the image of this point. Let's now begin with the solution. Let the image of point P having coordinates 3A on line mirror AB point Q having coordinates alpha beta. Q is the image of point P. Therefore, PQ is perpendicularly bisected at say point E. PQ therefore, we can find the coordinates of E by using the midpoint formula. So, coordinates of point ER as 3 by 2 beta plus 8 by 2. E lies on AB. Therefore, it must satisfy the equation of AB. Thus, by substituting x as alpha plus 3 by 2 and y as beta plus 8 by 2, we get alpha plus 3 by 2 plus 3 into beta plus 8 by 2 is equal to 7. Now this implies alpha plus 3 plus 3 beta plus 24 is equal to Q and this implies alpha plus 3 beta is equal to minus 13. Let's name this equation as equation number one. Given equation of AB is x plus 3y is equal to 7. Now this implies 3y is equal to minus x plus 7 and this implies y is equal to minus 1 by 3 into x plus 7 by 3. And comparing this equation with y is equal to mx plus c, we find that slope of AB is minus 1 by 3. Now since AB is perpendicular to PQ of slopes of AB and PQ is equal to minus 1, slope of AB is minus 1 by 3 and slope of PQ is beta minus a upon alpha minus 3. Since slope of line passing through points x1, y1 and x2, y2 is y2 minus y1 upon x2 minus x1 and this product is equal to minus 1 because AB is perpendicular to PQ. Now this implies beta minus 8 is equal to 3 into alpha minus 3. This implies beta minus 8 is equal to 3 alpha minus 9. This implies 3 alpha minus beta is equal to 1. Now let's name this equation as equation number two. Now by solving equation one and two, we can get the coordinates of points Q. So let's now solve equation one and equation two. Equation one is alpha plus 3 beta is equal to minus 13 and equation two is 3 alpha minus beta is equal to 1. Let's multiply the second equation by 3 so we get. Now adding these two equations we get tan alpha is equal to minus 10 and this implies alpha is equal to minus 1. Now by substituting the value of alpha in first equation we can get the value of beta. So by substituting alpha is minus 1 we get minus 1 plus 3 beta is equal to minus 13. This implies 3 into beta is equal to minus 12 and this implies beta is equal to minus 4. Hence the limit of the point 3A with respect to the 9, x plus 3 by z equal to 7 is minus 1 minus 4. This is our required answer. So this completes the session. Bye and take care.