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The straight line part 2

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Parallel and perpendicular lines

Now increase your knowledge of the straight line basics. The following equations are more advanced, but are built on the facts and equations you covered in the previous section. For each, we'll work through an example and show you the thought processes and workings you should produce in the exam. Remember, always show your working to gain full marks!

Gradients of lines which are perpendicular to each other.

Consider two lines with equations y = m_1 x + 3 and y = m_2 x - 7. If the lines are perpendicular to each other then m_1 \times m_2 = - 1and if m_1 \times m_2 = - 1 then the lines are perpendicular to each other.

Show that the lines with equations 4x + 2y - 8 = 0 and 2y = x + 1 are perpendicular.

First rearrange each equation in the form y = mx + c

\eqalign{ 2y &=& x + 1 \cr \cr y &=& {1 \over 2}x + {1 \over 2} \cr}

Identify the first gradient:

gradient = {1 \over 2}

\displaylines{ 4x + 2y - 8 &=& 0 \cr \cr 2y &=& - 4x + 8 \cr \cr y &=& {{ - 4} \over 2}x + {8 \over 2} \cr \cr y &=& - 2x + 4 \cr}

Identify the second gradient:

            gradient = -2

Complete the proof:m_1 \times m_2 = {1 \over 2} \times (- 2) = - 1

Hence lines are perpendicular

Intersection of two lines

Two non-parallel lines will have a common point - the point of intersection!

Find the point of intersection of the lines 2x + y - 8 = 0 and 4x - 3y = 6

First rearrange one of the lines in the form y = or x =

\eqalign{ 2x + y - 8 &=& 0 \cr y &=& - 2x + 8 \cr}

Substitute this result into the other line

\eqalign{ 4x - 3y &=& 6 \cr 4x - 3( - 2x + 8) &=& 6 \cr 4x + 6x - 24 &=& 6 \cr 10x &=& 30 \cr x &=& 3 \cr}

 

Complete the solving process

\eqalign{ y &=& - 2x + 8 \cr &=& - 2 \times 3 + 8 \cr &=& 2 \cr}

 

Point of intersection is (3,2)

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