amitasagar23 wrote:
Line l is tangent to a circle whose centre is at (3, 2). If the point of tangency is (6, 6), what is the slope of line l ?
A. -4/3
B. -3/4
C. 0
D. 3/4
E. 4/3
Hi, can someone please help me explain the solution of the above question. According to the solution set that i have, the ans is A, but i am not sure how. Thank you.
There's more the one way to solve this question; however, within the scope of realistic GMAT practice- the most simple way is to just use the slope formula and take the negative reciprocal. Why? What coordinates should be used for the slope formula? Firstly, this question tells us that the point of tangency lies on the coordinate (6,6) - what this simply means is that at this coordinate is a point that part of both on the circle AND on the line that touches it. If we plug both coordinates into the slope equation
y2-y1/ x2-x1= this would equal the slope of the circle's radius
According to geometric principles, if we take the negative reciprocal of the slope of the radius just like how we take the negative reciprocal of a parallel line to find the slope of its perpendicular line then we would have the line of the tangent line.
Hence B