Moderate -0.5 This is a straightforward coordinate geometry question requiring standard techniques: find midpoint, calculate gradient, use perpendicular gradient property, form equation of line, then substitute y=0. All steps are routine with no problem-solving insight needed, making it slightly easier than average but not trivial since it requires multiple coordinated steps.
(2, 7) to (10, 3) Mid-point (6, 5) Gradient = −½ Perp gradient = 2 Eqn \(y−5=2(x−6)\) Sets y to 0, → \((3\frac{1}{2}, 0)\)
B1 B1 B1✓ M1 A1 [5]
co co co Must be correct form of Perp co \(x = 3\frac{1}{2}\) only is ok.
(2, 7) to (10, 3) Mid-point (6, 5) Gradient = −½ Perp gradient = 2 Eqn $y−5=2(x−6)$ Sets y to 0, → $(3\frac{1}{2}, 0)$ | B1 B1 B1✓ M1 A1 [5] | co co co Must be correct form of Perp co $x = 3\frac{1}{2}$ only is ok.
1 Find the coordinates of the point at which the perpendicular bisector of the line joining (2, 7) to $( 10,3 )$ meets the $x$-axis.
\hfill \mbox{\textit{CAIE P1 2014 Q1 [5]}}