# Linear equation - Graphing - Cant find 2nd point

Printable View

• Mar 5th 2011, 07:32 PM
Dandonny
Linear equation - Graphing - Cant find 2nd point
Hey,

My problem is i have one point (2.15,1.1) in meters.
ive been told i want to find the other point/coordinate but i am unsure how to do so.
I know the gradient of my linear line is -3.7/1.8 and the distance away from point 1 is going to be 1.2metres.

I am unsure what to do, i have found the equation of this line which is Y=(-3.7/1.8)x + 5.52.

Thanks in advance to anybody who knows what i should do
• Mar 5th 2011, 10:59 PM
earboth
Quote:

Originally Posted by Dandonny
Hey,

My problem is i have one point (2.15,1.1) in meters.
ive been told i want to find the other point/coordinate but i am unsure how to do so.
I know the gradient of my linear line is -3.7/1.8 and the distance away from point 1 is going to be 1.2metres.

I am unsure what to do, i have found the equation of this line which is Y=(-3.7/1.8)x + 5.52.

Thanks in advance to anybody who knows what i should do

1. To be exact (as much as possible) your equation of the line is:

$y = -\dfrac{37}{18} x +\dfrac{1987}{360}$

2. The point you are looking for has the coordinates $P(p,y_P)$ with $y_P= -\dfrac{37}{18} p +\dfrac{1987}{360}$

3. Now use the distance formula which yields the distance between 2 points:

$d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

With your example you'll get:

$1.2=\sqrt{(p-2.1)^2+\left(-\dfrac{37}{18} p +\dfrac{1987}{360}-1.1\right)^2}$

4. Solve for p and consequently calculate $y_P$. You should get two different points!