Question

State whether the two lines through (6,3) and (1,1) and through (-2,5) and (2,-5) are parallel, perpendicular or neither.

Answer 1

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 answer:
Finding the slope of the line joining (6,3) and (1,1):
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 ${{x}_{1}}=6,{{x}_{2}}=1,{{y}_{1}}=3$ and ${{y}_{2}}=1$
Hence the slope of the line is $m=\dfrac{1-3}{1-6}=\dfrac{-2}{-5}=\dfrac{2}{5}$
Finding the slope of the line joining (-2,5) and (2,-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 ${{x}_{1}}=-2,{{x}_{2}}=2,{{y}_{1}}=5$ and ${{y}_{2}}=-5$
Hence the slope of the line is $m=\dfrac{-5-5}{2-\left( -2 \right)}=\dfrac{-10}{4}=\dfrac{-5}{2}$
Product of slope of the lines $=\dfrac{2}{5}\times \dfrac{-5}{2}=-1$
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 (6,3), we have
$6m+c=3$
Also, since the line passes through (1,1), we have
$m+c=1$
Hence, we have
$6m-m=3-1\Rightarrow m=\dfrac{2}{5}$
Hence the slope of AB is $\dfrac{2}{5}$
Let the equation of CD be y = mx+c
Since the line passes through (-2,5), we have
$-2m+c=5$
Also, since the line passes through (2,-5), we have
$2m+c=-5$
Hence, we have
$2m+2m=-5-5\Rightarrow m=\dfrac{-5}{2}$
Hence the slope of CD is $\dfrac{-5}{2}$
Hence the lines are perpendicular to each other.