 Hello friends, let's discuss the following question. It says, find the equation of a line drawn perpendicular to the line x upon 4 plus y upon 6 is equal to 1 through the point where it needs y x's. For this question we need to know the equation of line passing through the point say x1, y1 and having slope m and its equation is given by y minus y1 is equal to m into x minus x1. So this knowledge will work as key idea for this question. Let us now proceed on with the solution where given that the required line is perpendicular to the line x upon 4 plus y upon 6 is equal to 1 and in this line 4 is the x-intercept and 6 is the y-intercept. That means it cuts the x-axis at the point 40 and it cuts the y-axis at the point 06 and we have to find the equation of the line which is perpendicular to this line and where it needs the y-axis. That means the line passes through the point 06. Let's call this equation as 1. Now we simplify 1, take the same so the equation 1 becomes 6x plus 4y is equal to 24. This implies y is equal to minus 6x upon 4 plus 24 upon 4 and that means the slope of given line is 6 upon 4 which is equal to minus 3 by 2. Now we can find the slope of the line perpendicular to the given line that is slope of the line AC and we know that the slope of the line perpendicular to the given line is equal to minus 1 upon the slope of the given line. So this is equal to minus 1 upon minus 3 upon 2 and this is equal to 2 upon 3. Now we know that the required line passes through the point 06 and having slope 2 upon 3. So equation of line passing through 06 and having slope 3 is given by y minus 6 is equal to 2 upon 3 into x minus 0. Using the key idea this implies 2y minus 6 is equal to 2x and this implies 2x minus 3y plus 18 is equal to 0. Hence the required line is 2ax minus 3y plus 18 is equal to 0. So this completes the question. Hope you enjoyed the session. Goodbye and take care.