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Functions, 2ADV EQ-Bank 12

  1. Show the equation of \(AC\) is \(3 x+4 y-16=0\).   ( 2 marks)

  2. Show that \(BC\) is not perpendicular to \(AC\).   (2 marks)

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a.   \(\text{Proof (See Worked Solutions)}\)

b.   \(\text{Proof (See Worked Solutions)}\)

Show Worked Solution

a.    \(A(0,4), C(4,1)\)

\(m_{AC}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{1-4}{4-0}=-\dfrac{3}{4}\)

\(\text{Equation of line with} \ \ m=-\dfrac{3}{4} \ \ \text{through}\ (0,4):\)

\(y-y_1\) \(=m\left(x-x_1\right)\)
\(y-4\) \(=-\dfrac{3}{4}(x-0)\)
\(y-4\) \(=-\dfrac{3}{4} x\)
\(4 y-16\) \(=-3 x\)
\(3 x+4 y-16\) \(=0\)

 

b.    \(B(3,0), C(4,1)\)

\(m_{BC}=\dfrac{1-0}{4-3}=1\)

\(m_{AC} \times m_{BC}=-\dfrac{4}{3} \times 1=-\dfrac{4}{3} \neq-1\)

\(\therefore AC \ \text{is not} \ \perp \text{to} \ BC.\)

Functions, 2ADV EQ-Bank 2

  1. Find the equation of the line that passes through \((2,1)\) and \((-3,4)\).   (2 marks)

  2. Determine whether \((7,-2)\) lies on the line.   (1 mark)

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a.    \(y=\dfrac{4-1}{-3-2}=-\dfrac{3}{5}\)

b.    \(\text {Substitute}\ (7,-2) \ \text{into equation:}\)

\(-2\) \(=-\dfrac{3}{5} \times 7+\dfrac{11}{5}\)
\(-2\) \(=-\dfrac{21}{5}+\dfrac{11}{5}\)
\(-2\) \(=-2 \ \text{(correct)}\)

 

\(\therefore (7,-2) \text{ lies on line.}\)

Show Worked Solution

a.    \((2,1),(-3,4)\)

\(\text{Gradient}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{4-1}{-3-2}=-\dfrac{3}{5}\)

\(\text{Find equation with} \ \ m=-\dfrac{3}{5} \ \ \text{through}\ \ (2,1):\)

\(y-1\) \(=-\dfrac{3}{5}(x-2)\)
\(y\) \(=-\dfrac{3}{5} x+\dfrac{11}{5}\)

 

b.    \(\text {Substitute}\ (7,-2)\ \text{into equation:}\)

\(-2\) \(=-\dfrac{3}{5} \times 7+\dfrac{11}{5}\)
\(-2\) \(=-\dfrac{21}{5}+\dfrac{11}{5}\)
\(-2\) \(=-2 \ \text{(correct)}\)

 

\(\therefore (7,-2) \text{ lies on line.}\)

Functions, 2ADV F1 2024 HSC 1 MC

Consider the function shown.
 

Which of the following could be the equation of this function?

  1. \(y=2 x+3\)
  2. \(y=2 x-3\)
  3. \(y=-2 x+3\)
  4. \(y=-2 x-3\)
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\(C\)

Show Worked Solution

\(\text {Gradient is negative (top left } \rightarrow \text { bottom right)}\)

\(y \text{-intercept = 3 (only positive option)}\)

\(\Rightarrow C\)

Functions, 2ADV F1 2022 HSC 1 MC

Which of the following could be the graph of  `y= -2 x+2`?
 

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`A`

Show Worked Solution

`text{By elimination:}`

`y text{-intercept = 2  →  Eliminate}\ B and C`

`text{Gradient is negative  → Eliminate}\ D`

`=>A`

Functions, 2ADV F1 2016 HSC 12a*

The diagram shows points `A(1, 0), B(2, 4)` and `C(6, 1).` The point `D` lies on `BC` such that `AD _|_ BC.`
 

hsc-2016-12a
 

  1. Show that the equation of `BC` is  `3x + 4y-22 = 0`.   (2 marks)

  2. Determine the gradient of `AD`.   (1 mark)

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i.    `text(Proof)\ \ text{(See Worked Solutions)}`

ii.   `m_(AD)=4/3`

Show Worked Solution

i.    `B (2, 4),\ \ C (6, 1)`

`m_(BC) = (y_2-y_1)/(x_2-x_1) = (1-4)/(6-2) =-3/4`
 

`text(Equation of)\ \ BC,\ \ m=-3/4\ \ text(through)\ \ (2, 4):`

`y-y_1` `= m(x-x_1)`
`y-4` `=-3/4 (x-2)`
`4y-16` `= -3x + 6`
`3x + 4y-22` `= 0\ text(… as required.)`

 
ii.   `\text{Perpendicular lines:}\ m_1 xx m_2 = -1`

`m_(BC) =-3/4\ \ =>\ \ m_(AD)=4/3\ (BC _|_ AD)`

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`

Functions, 2ADV F1 2009 HSC 5a

In the diagram, the points `A` and `C` lie on the `y`-axis and the point `B` lies on the `x`-axis. The line `AB` has equation  `y = sqrt3x-3`. The line `BC` is perpendicular to `AB`.
 

2009 5a 

  1. Find the equation of the line `BC`.    (2 marks)

  2. Find the area of the triangle `ABC`.    (2 marks)

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  1. `y=-1/sqrt3 x +1`
  2. `text(Area)\ Delta ABC = 2 sqrt 3\ text(u²)`
Show Worked Solution

i.   `text(Gradient of)\ \ AB = sqrt 3`

`:. m_(BC) = -1/(sqrt3)\ \ (BC _|_ AB)`

`text(Finding)\ B,`

`0` `= sqrt3 x-3`
`sqrt 3 x` `= 3`
`x` `= 3/sqrt3 xx sqrt3/sqrt3= sqrt3`

 
`:. B (sqrt3, 0)`

 
`text(Equation of)\ \ BC\ \ text(has)\ \ m =-1/sqrt3\ \ text(through)\ \ (sqrt3, 0):`

`y\-y_1` `= m (x\-x_1)`
`y\-0` `=-1/sqrt3 (x\-sqrt3)`
`y` `=-1/sqrt3 x +1`

 

ii.  `AB\ \ text(cuts)\ y text(-axis when)\ \ x = 0, \ \ y=-3`

  `=> A (0,–3)`

`BC\ \ text(cuts)\ y text(-axis when)\ \ x = 0, \ \ y=1`

  `=> C (0,1)`

`:. AC` `= 4`
`OB` `= sqrt 3`

 

`text(Area)\ \ Delta ABC` `= 1/2 xx AC xx OB`
  `= 1/2 xx 4 xx sqrt 3`
  `= 2 sqrt 3\ text(u²)`