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Functions, 2ADV F1 2007 HSC 1f

Find the equation of the line that passes through the point `(1, 3)` and is perpendicular to  `2x + y + 4 = 0`.  (2 marks)

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`x-2y + 7 = 0`

Show Worked Solution
`2x + y + 4` `= 0`
`y` `= -2x-4`

  
`=>\ text(Gradient) = -2`

`:. text(⊥ gradient) = 1/2\ \ \ (m_1 m_2=-1)`
 

`text(Equation of line)\ \ m = 1/2, \ text(through)\ (1, 3):`

`y-y_1` `= m (x-x_1)`
`y-3` `= 1/2 (x-1)`
`y` `= 1/2 x + 5/2`
`2y` `= x + 5`
`:. x-2y + 5` `= 0`

Functions, 2ADV F1 2014 HSC 5 MC

Which equation represents the line perpendicular to  `2x-3y = 8`, passing through the point  `(2, 0)`?

  1. `3x + 2y = 4`
  2. `3x + 2y = 6`
  3. `3x-2y = -4`
  4. `3x-2y = 6`
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`B`

Show Worked Solution
`2x-3y` `= 8`
`3y` `= 2x-8`
`y` `= 2/3x-8/3`
`m` `= 2/3`
`:.\ m_text(perp)` `= -3/2\ \ \ (m_1 m_2=-1\text( for)_|_text{lines)}`

 

`text(Equation of line)\ \ m = -3/2\ \ text(through)\ \ (2,0):`

`y-y_1` `= m (x-x_1)`
`y-0` `= -3/2 (x-2)`
`y` `= -3/2x + 3`
`2y` `= -3x + 6`
`3x + 2y` `= 6`

 
`=>  B`