The two points (4,9) and (2,3) are on line A. A second line, line B is perpendicular to line A and goes through the point (2,3). What is the equation of line B?

Step 1, find the gradient of line A: gradient = change in y/change in x = (y1 - y2) / (x1 - x2) = 9-3/4-2 = 6/2 = 3Step 2, find the gradient of line B: as the two lines are perpendicular, the gradient of line B is the negative reciprocal of line A. Therefore you must multiply the gradient of line A by -1 then divide 1 by this value. i.e. if gradient of A = m then gradient of B = -1/m. So as gradient of line A is 3 , the gradient of line B is -1/3.Step 3, find the y-intercept of line B: the general equation of a line is y = mx + c and the gradient of line B is -1/3 so general equation is now y = -1/3x + c. From the question, we know line B goes through the point (2,3) so you can substitute these points into the line equation to find out c (y-intercept). y = 3 and x =2 so y = -1/3x + c is now 3 = (-1/3*2) + c, 3 = -2/3 + c, 11/3 = cOverall: y = -1/3x + 11/3