SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Functions, 2ADV F1 EQ-Bank 25

Prove that the line between \((1,-1)\) and \((4,-3)\) is perpendicular to the line \(3x-2y-4=0\)   (2 marks)

Show Answers Only

\(\text {Perpendicular lines}\ \ \Rightarrow\ m_1 \times m_2 = -1\)

\(\text {Line 1 gradient:}\)

\(P_1 (1,-1), P_2(4,-3) \)

\(m_1=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-3+1}{4-1}=-\dfrac{2}{3}\)
 

\(\text {Line 2 gradient:}\)

\(3x-2y-4=0\ \ \Rightarrow \ y= \dfrac{3}{2}x-2\ \ \Rightarrow m_2=\dfrac{3}{2}\)

\(m_1 \times m_2 = -\dfrac{2}{3} \times \dfrac{3}{2} = -1\)

\(\therefore\ \text{Lines are perpendicular.}\)

Show Worked Solution

\(\text {Perpendicular lines}\ \ \Rightarrow\ m_1 \times m_2 = -1\)

\(\text {Line 1 gradient:}\)

\(P_1 (1,-1), P_2(4,-3) \)

\(m_1=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-3+1}{4-1}=-\dfrac{2}{3}\)
 

\(\text {Line 2 gradient:}\)

\(3x-2y-4=0\ \ \Rightarrow \ y= \dfrac{3}{2}x-2\ \ \Rightarrow m_2=\dfrac{3}{2}\)

\(m_1 \times m_2 = -\dfrac{2}{3} \times \dfrac{3}{2} = -1\)

\(\therefore\ \text{Lines are perpendicular.}\)

EXAMCOPY Functions, 2ADV F1 2009 HSC 1a

Sketch the graph of  `y-2x = 3`, showing the intercepts on both axes.   (2 marks)

Show Answers Only

 

Show Worked Solution

`y-2x=3\ \ =>\ \ y=2x+3`

`ytext{-intercept}\ = 3`

`text{Find}\ x\ text{when}\ y=0:`

`0-2x=3\ \ =>\ \ x=-3/2`
 

Algebra, STD2 A4 SM-Bank 6 MC

A computer application was used to draw the graphs of the equations

`x - y = 4`  and  `x + y = 4`

Part of the screen is shown.

What is the solution when the equations are solved simultaneously?

  1. `x = 4, y = 4`
  2. `x = 4, y = 0`
  3. `x = 0, y = 4`
  4. `x = 0, y = −4`
Show Answers Only

`B`

Show Worked Solution

`text(Solution occurs at the intersection of the two lines.)`

`=> B`

Functions, 2ADV F1 2018 HSC 3 MC

What is the `x`-intercept of the line  `x + 3y + 6 = 0`?

  1. `(-6, 0)`
  2. `(6, 0)`
  3. `(0, -2)`
  4. `(0, 2)`
Show Answers Only

`A`

Show Worked Solution

`x text(-intercept occurs when)\ y = 0:`

`x + 0 + 6` `= 0`
`x` `= -6`

 
`:. x text{-intercept is}\  (-6, 0)`

`=>  A`

Functions, 2ADV F1 2017 HSC 1 MC

What is the gradient of the line  `2x + 3y + 4 = 0`?

  1. `-2/3`
  2. `2/3`
  3. `-3/2`
  4. `3/2`
Show Answers Only

`A`

Show Worked Solution
`2x + 3y + 4` `= 0`
`3y` `= -2x-4`
`y` `= -2/3 x-4/3`
`:.\ text(Gradient)` `= -2/3`

 
`=>  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)

NB. Parts ii-iii are not in the new syllabus.

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

a.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.)`

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)

Show Answers Only

`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 + 7/2`
`2y` `= x + 7`
`:. x-2y + 7` `= 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`
Show Answers Only

`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)

Show Answers Only
  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)`

Functions, 2ADV F1 2009 HSC 1a

Sketch the graph of  `y-2x = 3`, showing the intercepts on both axes.   (2 marks)

Show Answers Only

 

Show Worked Solution

`y-2x=3\ \ =>\ \ y=2x+3`

`ytext{-intercept}\ = 3`

`text{Find}\ x\ text{when}\ y=0:`

`0-2x=3\ \ =>\ \ x=-3/2`