Rakhi Mukherjee Sarkar
Last Activity: 14 Years ago
Q.1 Let the coordinates of B be (x1, y1) and that of C be (x2, y2). Therefore the mid point of BC is
(x1 + x2) /2 = 5 and (y1 + y2) /2 = 6 so, x1 = 10 - x2 & y1 = 12 - y2. But B(x1, y1) satisfies
2x + 3y = 29 & C(x2, y2) satisfies x + 2y = 16 therefore, 2x1 + 3y1 = 29__(i) & x2 + 2y2 = 16__(ii)
Putting the value of x1 & y1 in the first equation 7 then by simplifying we get, 2x2 + 3y2 = 27__(iii)
Now solving equation (ii) & (iii) we get, y2 = 3 and x2 = 10 so, equation passing through (5, 6) &
(10, 3) i.e BC is (x - 5) / (y - 6) = (5 - 10) / (6 - 3) i.e 3x + 5y = 45.