Question

Show that the three points $(4,2),(7,5)and(9,7)$ lie on a straight line.

Answer 1

Let the points be A(4, 2), B(7, 5) and C(9, 7). By the distance formula

$A{B}^2=(4-7{)}^2+(2-5{)}^2=(-3{)}^2+(-3{)}^2=9+9=18$

$B{C}^2=(9-7{)}^2+(7-5{)}^2=2^2+2^2=4+4=8$

$C{A}^2=(9-4{)}^2+(7-2{)}^2=5^2+5^2=25+25=50$

So, $AB=\sqrt{18}=\sqrt{9\times 2}=3\sqrt{2};BC=\sqrt{8}=\sqrt{4\times 2}=2\sqrt{2};CA=\sqrt{50}=\sqrt{25\times 2}=5\sqrt{2}$

This gives $AB+BC=3\sqrt{2}+2\sqrt{2}=5\sqrt{2}=AC$. Hence the points A, B and C are collinear.