Answer: True
Step-by-step explanation:
Let the point (7, 1) be Point A
Let the perpendicular bisector of the line joining (2,4) and (4,6) be Line PQ; P(2, 4) and Q(4, 6)
If A (7,1) lies on perpendicular bisector of P(2,4) and Q (4, – 6), then AP=AQ
[tex]\begin{aligned}\therefore \quad A P &=\sqrt{\left(2-7\right)^2+\left(4-1\right)^2}\:\\&=\sqrt{5^2+3^2}\\&=\sqrt{25+9}=\sqrt{34} \\\mathrm{and} \ \ \ A &=\sqrt{\left(4-7\right)^2+\left(6-1\right)^2}\: \\&=\sqrt{3^2+5^2} \\&=\sqrt{9+25}=\sqrt{34}\end{aligned}[/tex]
Therefore, (7, 1) does lie on the perpendicular bisector of the line joining (2,4) and (4,6)
