Plz Help Lads (1 Viewer)

[BzR_My]

The point P lies on the line f(x)=2x-9 and is equidistant from A and B, find the coordinates of P

 

[integral95]

What's A and B?

 

[BzR_My]

A(0,1) and B(2,3) Soz. And can anyone please write and explanation, would be highly appreciated.

 

[sy37]

graph the function, label points a,b, equidistant means they are in the middle of this a,b parameter.

 

[Drongoski]

The locus of points equidistant from points A and B is the perpendicular bisector of AB. This itself is a straight line, passing thru the mid-point of AB, viz: M([0+2]/2, [1+3]/2) = (1,2). Gradient of AB is (3-1)/(2-0) = 1; therefore gradient of the perp bisector is -1, with equation: y-2 = -1(x-1), i.e. y = -x + 3.

Now solve for the intersection of the 2 straight lines. y = -x +3 and y = 2x - 9. You should get x=4, y = -1.



Alternatively, you can make P(x,y) that lies on the given line and satisfying the condition: PA = PB.



edit
Why did you not supply the co-ords of A and B at the outset?

 

[Ambility]

Redacted.