Step-by-step explanation:
Given:
[tex] (x_1, \: y_1 ) =(2, \:3)\:\:\&\:\: (x_2, \: y_2) =(6, \:10)[/tex]
[tex]l(PQ) = \sqrt{ {(x_1 - x_2)}^{2} + {(y_1 - y_2)}^{2}} \\ \\ = \sqrt{ {(2 - 6)}^{2} + {(3 - 10)}^{2}} \\ \\ = \sqrt{ {( - 4)}^{2} + {( - 7)}^{2}} \\ \\ = \sqrt{16 + 49} \\ \\ = \sqrt{65} \\ \\ \therefore \purple{ \boxed{ l(PQ) = 8.06 \: units}}\\\\[/tex]
Let S be the mid-point of PQ.
[tex]\therefore S = \{\frac{x_1+x_2}{2}, \:\:\frac{y_1+y_2}{2}\} \\\\
= \{\frac{2 +6}{2}, \:\:\frac{3+10}{2}\} \\\\
= \{\frac{8}{2}, \:\:\frac{13}{2}\} \\\\
\therefore S= \{4, \:\:6.5\} \\\\[/tex]
Equation of line PQ is given as:
[tex] \frac{y-y_1}{y_1 - y_2}=\frac{x-x_1}{x_1 - x_2} \\\\
\therefore \frac{y-3}{3 - 10}=\frac{x-2}{2 - 6} \\\\
\therefore \frac{y-3}{-7}=\frac{x-2}{-4} \\\\
\therefore \frac{y-3}{7}=\frac{x-2}{4} \\\\
\therefore 4(y-3)=7(x-2)\\\\
\therefore 4y-12=7x-14\\\\
\therefore 7x-14 - 4y + 12=0\\\\
\red{\boxed {\therefore 7x-4y - 2 =0}} [/tex]
