1. ## quadratic equation

hi i'm struggling with this:

when a parabola crosses the x-axis at the points A and B, the axis of symmetry is the perpendicular bisector of the line joining the points A and B. since A is the point (-5, 0) and B is the point (2/3, 0) the equation of the line of symmetry is x = -13/6

i know that the mid point is -13/6 but why is that the equation? i thought you had to go through the whole y - y1 = m(x - x1) business. when the gradient of a horizontal line is 0 how do you come to the conclusion that the equation is x = -13/6.

thanks for any help

2. Originally Posted by mark
when a parabola crosses the x-axis at the points A and B, the axis of symmetry is the perpendicular bisector of the line joining the points A and B. since A is the point (-5, 0) and B is the point (2/3, 0) the equation of the line of symmetry is x = -13/6

i know that the mid point is -13/6 but why is that the equation?
Because the axis of symmetry is vertical. Vertical lines are $x=k$.

3. i'm sorry but i don't really understand what you mean, i literally took up maths about 2 weeks ago. what would the equation of the horizontal line be and how would you figure that out? to find the perpendicular bisector i know you need to reverse the gradient around ie m1 * m2 = -1 but i can't see how you'd do anything when the gradient is 0. why do you not use the y - y1 = m(x - x1) can you explain in detail please how and why its all done?

thanks

thanks

4. Originally Posted by mark
what would the equation of the horizontal line be and how would you figure that out?
Vertical lines have this form: $x=k$.
Horizontal lines have this form: $y=k$.

Examples.
A horizontal line through $(2,-3)$ is $y=-3$.
A vertical line through $(2,-3)$ is $x=2$.