 Hi, and welcome to the session. Let us discuss the following question. The question says, find the distance of the line 4x plus 7y plus 5 is equal to 0 from the point 1, 2 along the line 2x minus y is equal to 0. The AB real line whose equation is 4x plus 7y plus 5 is equal to 0. p be a point whose coordinates are 1, 2, and p2 be a line along this point whose equation is 2x minus y is equal to 0. We have to find pq. That is, we have to find distance of point p from this line. Let's now begin with the solution. Equation of line AB is 4x plus 7y is equal to 0. And equation of line pq is 2x minus y is equal to 0. Let's name this equation as equation number 1 and this as 2. Now, both these lines intersect at q, so we can find coordinates of q by solving these two equations simultaneously. So now, 2x minus y is equal to 0 implies y is equal to 2x. Now, replacing y by 2x in this equation, we get 4x plus 7 into 2x plus 5 is equal to 0. This implies 4x plus 14x plus 5 is equal to 0. And this implies x is equal to minus 5 by 18. From 2, we know that y is equal to 2x. So y is equal to 2 into minus 5 by 18. And this is equal to minus 5 by 9. So, coordinates of q are minus 5 by 18 and minus 5 by 9. Now, we can easily find distance between p and q by using the distance formula, right? Coordinates of p are 1, 2. And coordinates of q are minus 5 by 18, minus 5 by 9. Now, distance pq is equal to 1 plus 5 by 18 whole square plus 2 plus 5 by 9 whole square. This is equal to 23 by 18 whole square plus 23 by 9 whole square. Now, this is equal to 23 by 9 whole square into 1 by 2 whole square plus 1. This is equal to 23 by 9 into square root of 1 by 4 plus 1. And this is equal to 23 by 9 into root 5 by 2. And this is equal to 23 into root 5 by 18. Hence, the required distance is 23 into root 5 by 18 units. This is our required answer. So this completes the session. Bye and take care.