# Coordinate Geometry

#### Marshall

##### New member
There are two Points A(-1,0) and B(0,3).
(a) Find the gradient of AB. Answer: This is 3-0/0-(-1) = 3
(b) Find the equation of AB. Answer: y=3x+c, Therefore c = 3, and the equation is y=3x + 3
(c) If length of AB is √h, what is value of h. Answer:AB^2 =√(-1-0)^2 + (0-3)^2, so h=10
(d) If the point (-5,k) lies on the line produced, fine value of k. Answer, y=3(-5)+3 = -12. Therefore k is (-5, -12)
(e) If line y=x+1 is the line of symmetry of the triangle ABC, find the co-ordinates of C

#### pka

##### Elite Member
There are two Points A(-1,0) and B(0,3).
(a) Find the gradient of AB. Answer: This is 3-0/0-(-1) = 3
(b) Find the equation of AB. Answer: y=3x+c, Therefore c = 3, and the equation is y=3x + 3
(c) If length of AB is √h, what is value of h. Answer:AB^2 =√(-1-0)^2 + (0-3)^2, so h=10
(d) If the point (-5,k) lies on the line produced, fine value of k. Answer, y=3(-5)+3 = -12. Therefore k is (-5, -12)
(e) If line y=x+1 is the line of symmetry of the triangle ABC, find the co-ordinates of C

Note that $$\displaystyle A$$ is on the line $$\displaystyle y=x+1$$ so $$\displaystyle C$$ must be symmetric to $$\displaystyle B$$ with respect to that line.
I suggest you draw a picture (no graph paper if you have some). If you do not know about lines of symmetry look here.

#### Otis

##### Senior Member
... Therefore k is (-5, -12)
k is not an ordered pair of numbers; it's just -12

(e) If line y=x+1 is the line of symmetry of the triangle ABC, find the co-ordinates of C ...
I think they mean line y=x+1 bisects angle BAC.

If so, then the line passing through points B and C is perpendicular to line y=x+1 and the intersection point of those two lines is the midpoint of segment BC.

Write the equation of the line passing through points B and C, and then find the coordinates of the intersection point with line y=x+1 (that is, the midpoint coordinates of segment BC).

Let (h,k) be the coordinates of point C

You now have everything you need to use the formulas for the midpoint coordinates, to solve for h and k.

If you'd like more help, please show how far you got.

#### pka

##### Elite Member
I think they mean line y=x+1 bisects angle BAC.
The problem definitely stated that $$\displaystyle y=x+1$$ is an axis of symmetry.
As such, because $$\displaystyle A$$ is on that line then the line must be the bisector if $$\displaystyle \angle BAC$$.
The line $$\displaystyle y=-x+3$$ will contain both $$\displaystyle B~\&~C$$ and is perpendicular to $$\displaystyle y=x+1$$
Moreover, I think that the author wanted the point $$\displaystyle C$$ to be the perpendicular distance from the line as is $$\displaystyle B$$.

#### Otis

##### Senior Member
... problem definitely stated that $$\displaystyle y=x+1$$ is an axis of symmetry.
That's very possible; however, the op states that y=x+1 is "the" line of symmetry of a triangle. I'm not sure I've seen an axis of symmetry described like that. Glad to know I guessed correctly, heh.

... I think that the author wanted the point $$\displaystyle C$$ to be the perpendicular distance from the line as is $$\displaystyle B$$.
Yes, writing the distance formula and k=h+1 was my start, until I realized that midpoint formulas would not involve squared terms when solving the resulting system of two equations.

I also considered shifting the origin to (-1,0) and then viewing y=x+1 as y=x shifted. That way, AC would be the inverse of AB (reflected across the new y=x), so a run of 1 and a rise of 3 takes us from A to B and a run of 3 and a rise of 1 goes from A to C. (A better explanation of that would have been a lot more typing, but I'm still feeling wiped out from a trip to southern California.)

#### pka

##### Elite Member
That's very possible; however, the op states that y=x+1 is "the" line of symmetry of a triangle. Glad to know I guessed correctly, heh.
One cannot say the axis of symmetry because there can be as many as three. An equilateral triangle has three.