Co-ordinate geometry (1 Viewer)

[K@LLo]

hi i had trouble with these questions

1. A (0,0), B(2,1) and C(1,5) are the vertices of a triangle ABC. The triangle is rotated anti-clockwise abiut the point A, through a right angle in the X-Y plane. Find the equation if the image lines of:

a. the line AC
b. the line AB
c. the line BC

2. ABCD is a quadrilateral. The co-ordinates of A, B and C are (-8,6) (2,4) a.nd (5, -7) respectively. If the diagonals are perpendicular, and DC is parallel to the X-axis, find:

a. the co-ordinates of D.

3. The co-ordinates of the Vertices A,B,C of a triangle are (-1,3) (2,5) and (1,-1) respectively. Find:

a. the equation of the perpendicular line from A to BC
b. The equation of the line through B, parallel to this perpendicular line.

Thanks alot

 

[Aerath]

I wish I could scan you the answers, but I am capped. So you'll just have to make do with my shitty typed ones until someone else comes along.

Question 1

A(0,0) rotated through 90* = A'(0,0)
B(2,1) rotated through 90* = B'(1,-2)
C(1,5) rotated through 90* = C'(5,-1)

You can find these new points by using a diagram - and by inspection, the y value of the old point becomes the x value of the new point, and the x value of the old point is the y value of the new point, albeit, negative.

So to find the new line AC (which is A'C'), use point gradient formula:
m = (5-0) / (-1-0) = -5
y -0 = -5x - 0
y = -5x
Therefore A'C' = y = -5x

Similarly, with the other two, just find the gradients of the line, and use point gradient formula.

Question 2
I dunno. If diagonals are perpendicular, it means rhombus or square, but, I dunno how to make DC parallel to x axis. Sorry.

Question 3
a) First, find equation of the line travelling through BC and put in general form.
m = [5+1]/[2-1] = 6
y+1 = 6(x-1)
6x - y - 7 = 0
We know that A(-1,3)

Using perpendicular distance formula:
d = |Ax + By + C| / sqrt(A^2 + B^2)
= |[6(-1) - 3 - 7]|/sqrt37
= 16/sqrt37
= 16sqrt37 / 37

b) Since this line is parallel to BC, and BC has a gradient of 6 - this new line will have a gradient of -1/6.
Since it travels through B(2,5), we can use point gradient formula:
y - 5 = -1/6 (x-2)
6y - 30 = -x + 2
x + 6y - 32 = 0



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Yeah, no guarantees on the answer being right.

 

[lyounamu]

hi i had trouble with these questions

2. ABCD is a quadrilateral. The co-ordinates of A, B and C are (-8,6) (2,4) a.nd (5, -7) respectively. If the diagonals are perpendicular, and DC is parallel to the X-axis, find:

a. the co-ordinates of D.

Gradient of AC = -7-6/(5--8) = -1
So the gradient of BD = 1 since it is prependicular.

m = 1, (2,4)
y - 4 = x - 2
y = x + 2

Since C lies on DC, its y-value must be -7.
So -7 = x + 2
x = -9

C is (-9, - 7).

 

[Aerath]

Hmmmm, seems so simple. =\