Need Help with this quadratics stuff

**scubasteve94** · February 14th 2009, 10:11 PM

Diagram attatched
Find:
i) the equation of the line through points A and D

ii) the coordinates of D

iii) the equation of the perpendicular bisector of AB

iv) the coordinates of C

v) the area of triangle ADC

vi) the area of the quadrilateral ABCD

P.S Angle BAD is a right angle and C lies on the perpendicular bisector of AB. The equation of the line through points B and C is 3y=4x-14
Please show working!!
You dont have to answer all question any would be nice. Thanks

**Grandad** · February 14th 2009, 10:42 PM

Hello scubasteve94

Originally Posted by scubasteve94

Diagram attatched
Find:
i) the equation of the line through points A and D

ii) the coordinates of D

iii) the equation of the perpendicular bisector of AB

iv) the coordinates of C

v) the area of triangle ADC

vi) the area of the quadrilateral ABCD

P.S Angle BAD is a right angle and C lies on the perpendicular bisector of AB. The equation of the line through points B and C is 3y=4x-14
Please show working!!
You dont have to answer all question any would be nice. Thanks

Not quite sure why you've called it Quadratics, but here goes:

(i) Gradient of AB = $\frac{6-8}{8-2} = -\tfrac{1}{6}$

$\Rightarrow$ gradient of AD = 6

$\Rightarrow$ equation of AD is $y-8=6(x-2)$

i.e. $y = 6x-4$

(ii) D is intercept of this line with y-axis; i.e. (0, -4)

(iii) Perpendicular bisector of AB also has gradient 6, and passes through the mid-point of AB, i.e. (5, 7). So its equation is

$y -7=6(x-5)$

i.e. $y = 6x -23$

(iv) The line in (iii) meets AC at C, where the equations are simultaneously true; i.e. where

$3(6x-23)=4x-14$

$\Rightarrow 18x - 69 = 4x-14$

$\Rightarrow 14x = 55$

$\Rightarrow x=\frac{55}{14}$

I have to go now. Perhaps someone else can pick this up?

Grandad

**earboth** · February 14th 2009, 11:19 PM

Originally Posted by scubasteve94

i need to find the equation of the line through points A and D so..
...

I need some more information, otherwise I can't help you:

1. Which kind of quadrilateral do you have. (According to your sketch it could be a rectangle(?))

2. Is the point D placed on the y-axis?

**scubasteve94** · February 14th 2009, 11:26 PM

It does not specify what kind of quadrilateral it is but i dont think it is a rectangle.

Yes D is on the y axis

**earboth** · February 14th 2009, 11:28 PM

Originally Posted by scubasteve94

It does not specify what kind of quadrilateral it is but i dont think it is a rectangle.

Yes D is on the y axis

The slope of the line passing through A and B is:

$m_{AB}=\dfrac{6-8}{8-2}=-\dfrac13$

Since $AB \perp AD$ the slope of the line AD is:

$m_{AD}=-\dfrac1{m_{AB}}=3$

Now use the point-slope-formula of a straight line to determine the equation of the line AD:

$y-8=3(x-2)~\implies~\boxed{y=3x+2}$

And now you know that D(0, 2)

**scubasteve94** · February 14th 2009, 11:49 PM

I now have the equation of the perpendicular bisector of AB but i am unsure how to use this to find the coordinates of C. Grandad subbed it into the equation of BC whis is 3y=4x-14 but i am unsure why he did this?

**earboth** · February 14th 2009, 11:52 PM

Originally Posted by Grandad

[SIZE=3]Hello scubasteve94Not quite sure why you've called it Quadratics, but here goes:

(i) Gradient of AB = $\frac{6-8}{8-2} =\bold{\color{red} -\tfrac{1}{6}}$

...

I don't want to pick at you but ...

**earboth** · February 15th 2009, 12:00 AM

Originally Posted by scubasteve94

It does not specify what kind of quadrilateral it is but i dont think it is a rectangle.

Yes D is on the y axis

If the quadrilateral is actually a rectangle then the midpoint of BD must be the midpoint of AC too:

$M_{BD}\left(\frac{6+0}2\ ,\ \frac{8+2}2\right)$ that means: $M_{BD}\left(3, 5\right)$

$M_{BD} = M_{AC}\left(\frac{x_C+2}2\ ,\ \frac{y_C+8}2\right)$

Therefore $x_C=4, y_C=2$ that means the point C is $C(4, 2)$

**scubasteve94** · February 15th 2009, 12:02 AM

i no i also picked up on that and have made the correction to my notes!!
but im still unsure how to find the coordinates of C

**scubasteve94** · February 15th 2009, 12:06 AM

Originally Posted by earboth

If the quadrilateral is actually a rectangle then the midpoint of BD must be the midpoint of AC too:

yes that is true but it does not specify that it is a rectangle so..

**earboth** · February 15th 2009, 12:46 AM

Originally Posted by scubasteve94

i no i also picked up on that and have made the correction to my notes!!
but im still unsure how to find the coordinates of C

1. Calculate the equation of the perpendicular bisector of AB:
$<br /> M_{AB}(5, 7)$ Slog of the perpendicular bisector is m = 3 (compare my previous post)

Then $b: y - 7 = 3(x-5)~\implies~\boxed{y=3x-8}$

C is situated on the bisector and the line $BC: 3y=4x-14$

Calculate the intersection point:

$3(3x-8)=4x-14~\implies~5x=10~\implies~x=2$ . Plug in this value into the first equation: $y = 3\cdot 2 - 8 = -2$

Thus C(2, -2)

**Grandad** · February 15th 2009, 07:45 AM

Hello again -

Sorry for my silly error earlier. Of course 6 - 8 = -2, not -1.

(While I was teaching, I always used to tell my students that 99% of computational errors were either due to a minus sign or a factor of 2. So it's a factor of 2, then, this time!)

Anyway, let's complete this answer now.

Of course, as earboth has shown, D is (0, 2) and C is (2, -2). So the diagram is like the one I have attached. So:

(v) The area of triangle ADC $= \tfrac{1}{2}AC\cdot DE = \tfrac{1}{2}10\cdot 2 = 10$ sq units.

(vi) The area of triangle ABC = $\tfrac{1}{2}AC\cdot BF = \tfrac{1}{2}10\cdot 6 = 30$ sq units. So area of quadrilateral ABCD = 40 sq units.

Grandad

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