Math Help - coordinate geometry problem

1. Hi all,

1. Find $h$ if the point $(h,0)$ is equidistant from the points $(-1,7)$ and $(-3,-2)$

thanks

i tried using the perpendicular but this seems to be flawed.

$y=mx+c$

and

$y=-\frac{1}{m}x+c$

$m=\frac{y{2}-y{1}}{x{2}-x{1}}$

the functions were then

$y=\frac{9}{2}x+\frac{23}{2}$

and

$y=-\frac{2}{9}x+\frac{23}{2}$

therefore when y=0

$\frac{2x}{9}=\frac{23}{2}\equiv 4x=207$

$h=\frac{207}{4}$

should i be using the midpoint or something?

2. 1. Find $h$ if the point $(h,0)$ is equidistant from the points $(-1,7)$ and $(-3,-2)$

sqrt((h+1)^2+(0-7)^2)=sqrt((h+3)^2+(0+2)^2)
h=37/4

3. Thanks for the answer learnmath

i understand what you did, but have trouble visualising it!

4. Originally Posted by sammy28
i tried using the perpendicular but this seems to be flawed.

$y=mx+c$

and

$y=-\frac{1}{m}x+c$
.....
...

should i be using the midpoint or something?

Why do you think that the value of "c" will remain same.

________________________________________
Visualizing it

• Two points are given
• Make a line joining these points.
• Find mid-point of the given points in the line
• Draw a line perpendicular to previous line and passing through midpoint
• The second line will intersect with x-axis at some point. The x coordinate of this point is value of "h"

5. Hope this will help.

6. Thanks ADARSH and learnmath, i got it in the end

im sure i was supposed to answer this your way learnmath, ie using pythagoras. the drawing does help a lot.

midpoint is $\left(-\frac{4}{2},\frac{5}{2}\right)$

perpendicular line is

$y=-\frac{2}{9}x+c$

solving for $c$ by using midpoint gives our perpendicular line equation

$y=-\frac{2}{9}x+\frac{37}{18}$

now find the x intercept

$\frac{2x}{9}=\frac{37}{18} \equiv 36x=333$

therefore

$x=\frac{37}{4}$