 Hello and welcome to the session. In this session we discussed the following question that says, what is the image of the point 452 in the line x plus 1 upon 2 equal to y upon 3 equal to z minus 1 upon 4. Before we move on to the solution, let's recall one fact according to which we have that the two lines with direction ratios a1, b1, c1 and a2, b2, c2 are said to be perpendicular if a1, a2 plus b1, b2 plus c1, c2 is equal to 0. This is the key idea that we use for this question. Now we move on to the solution. We are given the line x plus 1 upon 2 is equal to y upon 3 is equal to z minus 1 upon 4. This is the line. Let this be equal to lambda and we take this as equation 1. We are given a point, say point p with coordinates 452 and we have to find the image of this point p on the line 1. So this is the given line 1. This is the point p. We have to find the image of this point on this line. We have drawn a perpendicular from the point p to the given line and let this point n be the foot of the perpendicular drawn from the point p with coordinates 452 to the line 1. So now from equation 1 we have x is equal to 2 lambda minus 1, y is equal to 3 lambda, z is equal to 4 lambda plus 1. So we have that n is the foot of the perpendicular drawn from the point p with coordinates 452 to the given line and we get that the coordinates of the point n are given as 2 lambda minus 1, 3 lambda and 4 lambda plus 1. So this is the point n with coordinates 2 lambda minus 1, 3 lambda and 4 lambda minus 1, 4 lambda plus 1. So now we get the direction ratios of p n are given as 2 lambda minus 1 minus 4, 3 lambda minus 5, 4 lambda plus 1 minus 2. So we get that the direction ratios of p n are 2 lambda minus 5, 3 lambda minus 5, 4 lambda minus 1 and from equation 1 the direction ratios of the given line 2, 3 and 4. Now since we have that p n perpendicular to the given line therefore using the results stated in the key idea we get minus 5 multiplied by 2 plus 3 lambda minus 5 multiplied by 3 plus 4 lambda minus 1 multiplied by 4 is equal to 0. So this gives us 4 lambda minus 10 plus 9 lambda minus 15 plus 16 lambda minus 4 is equal to 0. So we get 29 lambda minus 29 is equal to 0 that is 29 lambda is equal to 29 so we get lambda is equal to 1. Now that we have got the value for lambda so we get the coordinates of the point n are putting the value for lambda we get 2 into 1 minus 1 comma 3 into 1 comma 4 into 1 plus 1 that is the coordinates of point n are 2 minus 1, 1, 3 into 1, 3, 4 plus 1, 5. So 1, 3, 5 are the coordinates of the point n. So we have a point n with coordinates 1, 3, 5. We take this point m with coordinates alpha, beta, gamma to be the image of the point p in the given line that is let the point m with coordinates alpha, beta, gamma be the image of the point p with coordinates 4, 5, 2 in the given line then we say that this n is the midpoint of p n. So in that case we say 4 plus alpha upon 2 is equal to the x coordinate of the point n which is 1 then 5 plus beta upon 2 is equal to the y coordinate of the point n that is 3 then 2 plus gamma upon 2 is equal to the z coordinate of the point n which is 5. So from here we get alpha is equal to 2 minus 4 which is equal to minus 2 then beta is equal to 6 minus 5 which is equal to 1 and gamma is equal to 10 minus 2 equal to 8. We have got the values for alpha, beta and gamma thus we get the point m with coordinates minus 2, 1 and 8 and this point m is the image of the point p in the given line thus the image of the point p with coordinates 4, 5, 2 in the given line is point m with coordinates minus 2, 1, 8. So this is our final answer. This completes the session. Hope you have understood the solution of this question.