Finding the locus, seriously quick

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Homework Help Overview

The discussion revolves around finding the locus of a point P(x,y) given specific conditions related to two points A(1,4) and B(-3,2). The conditions include determining the locus where the angle APB is a right angle and where P is equidistant from point A and the x-axis.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using the concept of gradients to establish the condition for a right angle and the distance formula to find the locus equidistant from the x-axis. Questions arise regarding the understanding of the equations and the application of the gradient formula.

Discussion Status

The discussion is active, with participants seeking clarification on the mathematical concepts involved. Some guidance has been provided regarding the relationship between the gradients of perpendicular lines, and participants express appreciation for the assistance received.

Contextual Notes

There is an indication that the original poster is struggling with the mathematical concepts, particularly the application of the gradient formula and the implications of the conditions given for the locus.

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1. Given the points A(1,4) and B(-3,2), find the equation of each locus of the pont P(x,y), a) angle APB is a right angle
b) P is equidistant from A and the x-axu
 
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a) The key point is that it's a right angle, so simply use [tex]M_{PA}M_{PB}=-1[/tex] where M is the gradient of each line. You will end up with an equation in x and y, which is your locus.

b) Since P is equidistant from the x-axis (y=0) and A(1,4) use the distance formula.
 
thanks for your quick response, but could you please explain a) i seem to be having trouble i understanding the equation you have written.
 
Sure, no problem.

Are you familiar with the fact that if two lines are perpendicular to each other, their gradients multiply to give -1?

Such as y=x and y=-x, their gradients multiplied together give: (1)(-1)=-1 and thus these lines are perpendicular (right angles) to each other.

So [tex]M_{PA}[/tex] is the gradient of the line PA, which is found by using the gradient formula,
[tex]\frac{y_2-y_1}{x_2-x_1}[/tex]

If you're still unsure, just ask :smile:
 
Thanks! That makes it a lot more clearer now. Also thanks for your fast responses. =] Hope you have a great day =]
 
Glad I could help :smile: Have a great day too.
 

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