 Hello friends, let's discuss the following question. It says, find the equations of the lines which cut off intercepts on the x's whose sum and product are 1 and minus 6 respectively. Let us now proceed on with the solution. Now intercept form of line is given by x upon A plus y upon B is equal to 1 where A is the x-intercept and B is the y-intercept. Now we are given that sum of the intercepts is 1 and the product of intercepts is minus 6. So according to question A plus B is equal to 1 and A into B is equal to minus 6. A plus B is equal to 1 implies B is equal to 1 minus A. Substituting B is equal to 1 minus A in this we have A into 1 minus A is equal to minus 6 and this implies A minus A square is equal to minus 6 and this implies minus A square plus A plus 6 is equal to 0. Now taking minus A common we have A square minus A minus 6 is equal to 0. Now we will solve this quadratic equation for A. Let us factorize this quadratic equation so we have A square minus 3A plus 2A minus 6 is equal to 0. Now taking A common from first two terms we have A into A minus 3 and taking two common from last two terms we have 2 into A minus 3 is equal to 0. Now taking A minus 3 common we have A minus 3 into A plus 2 is equal to 0 and this implies A is equal to 3 or A is equal to minus 2. Now if A is equal to 3 then B is equal to 1 minus 3 that is minus 2 and if A is equal to minus 2 B is equal to 1 minus minus 2 that is 3. So we have obtained the intercepts as 3 minus 2 and minus 2 3. So the required equations minus y upon 2 is equal to 1 and x upon minus 2 plus y upon 3 is equal to 1. Now simplifying this equation taking LCM the first equation becomes 2x minus 3y is equal to 6 and simplifying second equation and taking LCM this equation becomes minus 3x plus 2y is equal to 6. This completes the question. Hope you enjoyed this session goodbye and take care.