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Line AB formed by (-3,8) and (3,2)Line CD formed by (7,1) and (5,-1)Parallel perpendicular or neither

User Shyam
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1 Answer

2 votes

For two lines to be parallel,


\begin{gathered} m_1=m_2 \\ m_1=Gradient\text{ of line AB} \\ m_2=\text{Gradient of line CD} \end{gathered}

For two lines to be perpendicular,


m_1* m_2=-1

The formula for the gradient of a line passing through points (x1,y1) and (x2,y2) is given as


m=(y_2-y_1)/(x_2-x_1)

For line AB, the point given is


\begin{gathered} (-3,8)\text{ and (3,2)} \\ x_1=-3,y_1=8,x_2=3,y_2=2 \end{gathered}

By substituting in the formula, we will have


\begin{gathered} m=(2-8)/(3-(-3)) \\ m=-(6)/(3+3) \\ m_1=-(6)/(6) \\ m_1=-1 \end{gathered}

For line CD, the point given is


\begin{gathered} (7,1)\text{ and (5,-1)} \\ x_1=7,y_1=1,x_2=5,y_2=-1 \end{gathered}

By substituting the values in the formula, we will have


\begin{gathered} m_2=(-1-1)/(5-7) \\ m_2=-(2)/(-2) \\ m_2=1 \end{gathered}
\begin{gathered} m_1\\e m_2 \\ \text{therefore the two lines are not parallel} \end{gathered}
\begin{gathered} m_1* m_2 \\ =-1*1 \\ =-1 \end{gathered}

Therefore,


\begin{gathered} \sin ce \\ m_1* m_2=-1 \\ \text{Then the two lines are perpendicular} \end{gathered}

Hence,

Line AB and line CD are perpendicular

User Laura Franco
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