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Question: State whether the two lines through (6,3) and (1,1) and through (-2,5) and (2,-5) are parallel, perp...

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

Explanation

Solution

Hint: Find the slope of the lines using the property that the slope of the line joining the points A(x1,y1)A\left( {{x}_{1}},{{y}_{1}} \right) and B(x2,y2)B\left( {{x}_{2}},{{y}_{2}} \right) is given by m=y2−y1x2−x1m=\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(x1,y1)A\left( {{x}_{1}},{{y}_{1}} \right) and B(x2,y2)B\left( {{x}_{2}},{{y}_{2}} \right) is given by m=y2−y1x2−x1m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}.
Here x1=6,x2=1,y1=3{{x}_{1}}=6,{{x}_{2}}=1,{{y}_{1}}=3 and y2=1{{y}_{2}}=1
Hence the slope of the line is m=1−31−6=−2−5=25m=\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(x1,y1)A\left( {{x}_{1}},{{y}_{1}} \right) and B(x2,y2)B\left( {{x}_{2}},{{y}_{2}} \right) is given by m=y2−y1x2−x1m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}.
Here x1=−2,x2=2,y1=5{{x}_{1}}=-2,{{x}_{2}}=2,{{y}_{1}}=5 and y2=−5{{y}_{2}}=-5
Hence the slope of the line is m=−5−52−(−2)=−104=−52m=\dfrac{-5-5}{2-\left( -2 \right)}=\dfrac{-10}{4}=\dfrac{-5}{2}
Product of slope of the lines =25×−52=−1=\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⊥CDAB\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=36m+c=3
Also, since the line passes through (1,1), we have
m+c=1m+c=1
Hence, we have
6m−m=3−1⇒m=256m-m=3-1\Rightarrow m=\dfrac{2}{5}
Hence the slope of AB is 25\dfrac{2}{5}
Let the equation of CD be y = mx+c
Since the line passes through (-2,5), we have
−2m+c=5-2m+c=5
Also, since the line passes through (2,-5), we have
2m+c=−52m+c=-5
Hence, we have
2m+2m=−5−5⇒m=−522m+2m=-5-5\Rightarrow m=\dfrac{-5}{2}
Hence the slope of CD is −52\dfrac{-5}{2}
Hence the lines are perpendicular to each other.