State whether the two lines through (3,4) and (6,2) and through (4,2) and (6,5) are parallel, perpendicular or neither.

Hint: Find the slope of the lines using the property that the slope of the line joining the points $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$. Use the fact that if the slopes of two lines are equal, then they are parallel to each other and if the product of the slopes of two lines is -1, then the lines are perpendicular. Hence determine whether the lines are parallel or perpendicular or neither.

Complete step-by-step solution:
Finding the slope of the line joining (3,4) and (6,2):
We know that the slope of the line joining the points $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$.
Here ${{x}_{1}}=3,{{x}_{2}}=6,{{y}_{1}}=4$ and ${{y}_{2}}=2$
Hence the slope of the line is $m=\dfrac{2-4}{6-3}=\dfrac{-2}{3}=\dfrac{-2}{3}$
Finding the slope of the line joining (4,2) and (6,5):
We know that the slope of the line joining the points $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$.
Here ${{x}_{1}}=4,{{x}_{2}}=6,{{y}_{1}}=2$ and ${{y}_{2}}=5$
Hence the slope of the line is $m=\dfrac{5-2}{6-\left( 4 \right)}=\dfrac{3}{2}=\dfrac{3}{2}$
Product of slope of the lines $=\dfrac{-2}{3}\times \dfrac{3}{2}=-1$
Now since the product of the slopes of the two lines is -1, the lines are perpendicular to each other.

Note: [i] Viewing graphically:

As is evident from the graph $AB\bot CD$
[ii] Alternative solution:
Let the equation of AB be y=mx+c
Since the line passes through (3,4), we have
$3m+c=4$
Also, since the line passes through (6,2), we have
$6m+c=2$
Hence, we have
$3m-6m=4-2\Rightarrow m=\dfrac{-2}{3}$
Hence the slope of AB is $\dfrac{-2}{3}$
Let the equation of CD be y = mx+c
Since the line passes through (4,2), we have
$4m+c=2$
Also, since the line passes through (6,5), we have
$6m+c=5$
Hence, we have
$6m-4m=5-2\Rightarrow m=\dfrac{3}{2}$
Hence the slope of CD is $\dfrac{3}{2}$
Hence the lines are perpendicular to each other.