Equation to find an unknown point on a graph

In summary, the location of point $E$ can be found using coordinate geometry, with coordinates $\left(\frac{b+c\cos\delta}{\sin\delta},c\right)$. The distance from $A$ to $E$ is given by $\frac{\sqrt{b^2+c^2 + 2bc\cos\delta}}{\sin\delta}$, and the angle that the line $AE$ makes with the horizontal is given by $\tan\theta = \frac{c\sin\delta}{b+c\cos\delta}$. I hope that helps!
  • #1
Jazoar
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Hi All,

I need help to find an equation that will give me the location of 'E' in the image below.

'A' is the starting point on a graph.

All values from 'B' to 'D' are known values.

Currently, 'B' = 50, 'C' = 20, & 'D' = 135° - but I need an equation that will find 'E' no matter what these values are.

View attachment 7931

Any help would be greatly appreciated!

- Jazoar
 

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  • #2
Jazoar said:
Hi All,

I need help to find an equation that will give me the location of 'E' in the image below.

'A' is the starting point on a graph.

All values from 'B' to 'D' are known values.

Currently, 'B' = 50, 'C' = 20, & 'D' = 135° - but I need an equation that will find 'E' no matter what these values are.
Any help would be greatly appreciated!

- Jazoar
Hi Jazoar, and welcome to MHB!

I found your diagram difficult to follow, because the letters A, B, C, D and E are being used to describe different things. If I understand it correctly, A and E are points, B and C are distances, and D is an angle. So I shall change the notation, keeping $A$ and $E$ as points but writing the distances as $b$ and $c$, and the angle as $\delta.$

I think the easiest way to solve the problem is to use coordinate geometry, taking $A$ as the origin, and the horizontal blue line as the $x$-axis. The horizontal red line then has the equation $y=c$, and the sloping red line has equation $-y\cos\delta + x\sin\delta = b.$ Those two lines meet at the point $E$, whose coordinates are $\left(\frac{b+c\cos\delta}{\sin\delta},c\right).$

From that, you can calculate that the distance from $A$ to $E$ is $$\frac{\sqrt{b^2+c^2 + 2bc\cos\delta}}{\sin\delta},$$ and if the line $AE$ makes an angle $\theta$ with the horizontal then $\theta$ is given by $$\tan\theta = \frac{c\sin\delta}{b+c\cos\delta}.$$ In the case when $b=50$, $c=20$ and $\delta = 135^\circ$, those formulas give $AE \approx 54.5$ and $\theta \approx 21.5^\circ.$
 
Last edited:

Related to Equation to find an unknown point on a graph

What is the equation to find an unknown point on a graph?

The equation to find an unknown point on a graph is called the slope-intercept form, which is y = mx + b. In this equation, m represents the slope of the line and b represents the y-intercept.

How do I use the equation to find an unknown point on a graph?

To use the equation, you need to know the value of the slope (m) and the y-intercept (b). Once you have these values, you can plug them into the equation and solve for y. This will give you the y-coordinate of the unknown point. Then, you can use the x-coordinate of the point to find the corresponding y-coordinate on the graph.

What if I don't know the values of the slope and y-intercept?

If you don't know the values of the slope and y-intercept, you can use other points on the graph to find them. You can choose two points on the graph and use the slope formula (m = (y2-y1)/(x2-x1)) to find the slope. Then, you can plug in one of the points and the slope into the equation y = mx + b to solve for b, the y-intercept.

Can I use this equation for any type of graph?

Yes, the slope-intercept form can be used for any type of graph, as long as the graph is a straight line. However, for non-linear graphs, you may need to use a different equation or method to find an unknown point.

Are there any limitations to using this equation?

While the slope-intercept form is a useful tool for finding an unknown point on a graph, it does have some limitations. It can only be used for linear graphs, and it assumes that the graph is accurate and the points are precise. Additionally, it may not give an exact solution if the graph is not perfectly linear.

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