Form an equation describe the curve? (1 Viewer)

[fabl]

The points A (4,-2) , B (-4,4) and P (x,y) form a right angle at P. Form an equation in x and y. hence find the equation of the curve on which P lies?

How to find the equation?

 

[QZP]

What's the condition for there to be a right angle at P?

 

[fabl]

Im thinking possibly perpendicular distance. however, didn't say on the question.

 

[Stygian]

PA and PB are perpendicular

meaning their gradients multiply to be -1

so find their gradients using the formulae y-y1/x-x1 and sub in the appropriate x1 and y1 values, equate the product of these gradients to -1 and proceed from there

(i meant y2-y1/x2-x1 but pre-emptively subbed in P(x,y) in lol)

 

[QZP]

Im thinking possibly perpendicular distance. however, didn't say on the question.

I was hinting at the answer. You have to look at the question and extract information (Stygian answered it- m(PA).m(PB) = -1)

 

[fabl]

Do you mean like this

(y +2) (y-4) / (x-4) (x+4) = -1

which leads to y-2y = -1x +24?

 

[Carrotsticks]

The points A (4,-2) , B (-4,4) and P (x,y) form a right angle at P. Form an equation in x and y. hence find the equation of the curve on which P lies?

How to find the equation?

A and B are the endpoints of the diameter of some circle, where P lies on its circumference.

So the centre of the circle is the midpoint of AB and the radius is half the distance AB. You can write down the equation of the circle from here.

 

[aDimitri]

A and B are the endpoints of the diameter of some circle, where P lies on its circumference.

So the centre of the circle is the midpoint of AB and the radius is half the distance AB. You can write down the equation of the circle from here.

but circle geo is 3U so the angle in a semi-circle theorem is outside of the scope of this course isn't it?

 

[Carrotsticks]

but circle geo is 3U so the angle in a semi-circle theorem is outside of the scope of this course isn't it?

Correct, didn't see that this was in the 2U section.

 

[fabl]

A and B are the endpoints of the diameter of some circle, where P lies on its circumference.

So the centre of the circle is the midpoint of AB and the radius is half the distance AB. You can write down the equation of the circle from here.

How did you figure out the curve was a circle?

 

[Carrotsticks]

How did you figure out the curve was a circle?

It's a Circle Geometry (not in the 2U course) theorem called Thales' Theorem.

Any angle subtended from the diameter of a circle onto the circumference is 90 degrees. Since angle APB is 90 degrees, A and B must be the endpoints of the diameter, and P must lie on the circumference.