Moderate -0.3 This is a straightforward coordinate geometry problem requiring simultaneous equations from two standard formulas (distance and gradient). Students apply the gradient formula to get b = 2a - 4, substitute into the distance formula to get a quadratic, then solve for both variables. While it involves multiple steps, the techniques are routine and the algebraic manipulation is standard for A-level, making it slightly easier than average.
7 The coordinates of points \(A\) and \(B\) are \(( a , 2 )\) and \(( 3 , b )\) respectively, where \(a\) and \(b\) are constants. The distance \(A B\) is \(\sqrt { } ( 125 )\) units and the gradient of the line \(A B\) is 2 . Find the possible values of \(a\) and of \(b\).
7 The coordinates of points $A$ and $B$ are $( a , 2 )$ and $( 3 , b )$ respectively, where $a$ and $b$ are constants. The distance $A B$ is $\sqrt { } ( 125 )$ units and the gradient of the line $A B$ is 2 . Find the possible values of $a$ and of $b$.
\hfill \mbox{\textit{CAIE P1 2014 Q7 [6]}}