Coordinate Geometry

**Ai Ekio** · July 8th 2009, 05:30 PM

The directions say I'm supposed to sketch each locus on a coordinate plane..so graph paper..
1. All points 3 units from line y=-2
2. All points equidistant from points (2,4) and (0,0)

I'm clueless as to what the resulting sketches are supposed to look like..where is 'line y=-2" on the coordinate plane anyway?Please help,I'm totally confused.Thank you..

**Shyam** · July 8th 2009, 06:15 PM

Originally Posted by Ai Ekio

The directions say I'm supposed to sketch each locus on a coordinate plane..so graph paper..
1. All points 3 units from line y=-2

All the points 3 units from the line y = - 2 lie on a line which is 3 units from the given line. So, the locus is the line y = 1 and line y = - 5

please see attached graph.

**Shyam** · July 8th 2009, 06:33 PM

Originally Posted by Ai Ekio

The directions say I'm supposed to sketch each locus on a coordinate plane..so graph paper..
2. All points equidistant from points (2,4) and (0,0)

The locus of the points equi-distant from these points, is the line which is perpendicular bisector of the line joining these two given points.

The midpoint between the points (2, 4) and (0,0) is $\left(\frac{2+0}{2}, \frac{4+0}{2}\right)=(1, 2)$

$Slope =\frac{4-0}{2-0}=2$

Slope of perpendicular line $= \frac{-1}{2}$

Equation of locus line (pink line in the graph) is

$y = \frac{-1}{2}x+b$

$2 = \frac{-1}{2}(1)+b$

$b = \frac{5}{2}$

equation of locus line is

$y = \frac{-1}{2}x+\frac{5}{2}$

$x + 2y = 5$

Thread: Coordinate Geometry

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