Y-Intercept of Crop Duster Flight Path

[Serious Max]

Homework Statement

A crop dusting airplane flying a constant speed of 120 mph is spotted 2 miles South and 1.5 miles East of the center of a circular irrigated field. The irrigated field has a radius of 1 mile. Impose a coordinate system as pictured, with the center of the field the origin (0,0). The flight path of the duster is a straight line passing over the labeled points P and Q. Assume that the point Q where the plane exits the airspace above the field is the Western-most location of the field. Answer these questions:

1. Find a linear equation whose graph is the line along
which the crop duster travels.

2. Relevant graph

The Attempt at a Solution

The problem is just an example given in the textbook, and here's their solution:

1. Take Q = (−1, 0) and S = (1.5,−2) = duster spotting point. Construct a line through Q and S. The slope is −0.8 = m and the line equation becomes:
y = −0.8x − 0.8And my question is where did this y = −0.8x − 0.8 come from? Is it y-intercept? I'd assume it should be maybe: y = −0.8x − 2

 

[SteamKing]

Homework Statement

A crop dusting airplane flying a constant speed of 120 mph is spotted 2 miles South and 1.5 miles East of the center of a circular irrigated field. The irrigated field has a radius of 1 mile. Impose a coordinate system as pictured, with the center of the field the origin (0,0). The flight path of the duster is a straight line passing over the labeled points P and Q. Assume that the point Q where the plane exits the airspace above the field is the Western-most location of the field. Answer these questions:

1. Find a linear equation whose graph is the line along
which the crop duster travels.

2. Relevant graph

The Attempt at a Solution

The problem is just an example given in the textbook, and here's their solution:

1. Take Q = (−1, 0) and S = (1.5,−2) = duster spotting point. Construct a line through Q and S. The slope is −0.8 = m and the line equation becomes:
y = −0.8x − 0.8


And my question is where did this y = −0.8x − 0.8 come from? Is it y-intercept? I'd assume it should be maybe: y = −0.8x − 2

What is the value of y in your equation when x = 0? Does this equation match then match the diagram?

Do you know how to find the y-intercept of a linear equation?

 

[LCKurtz]

@maxpancho: You have two points given on the line, $(-1,0)$ and $(\frac 3 2, -2)$. Do you know how to write the equation of a straight line through two given points? Try it and see what you get.

 

[Serious Max]

Well, okay. So it's like I find b by taking a point (0,b) and then plug it into the origial equation. But that's what initially confused me click ...that they used y1 as it is.

But I think I understand now. It's just another way of writing an equation of a line. In the first example it is written in a slope-intercept form, which is y = mx + b. And thus here we must first find the y-intercept m and then plug it back into the original equation.

And in the second example it is written in a point-slope formula y = m(x − x1) + y1. Because I guess it is more convenient in this case, since m here would equal −2168140,17647058... But the equation is still valid, just not very practical.

So I figure I could write the equation from the initial example as: y = −0.8(x − 1.5) − 2, which gives the same y = −0.8x −0.8 (oh wait, so then there was no need to plug 0,b trying to find m)...



Is it correct?

 

[haruspex]

Well, okay. So it's like I find b by taking a point (0,b) and then plug it into the origial equation. But that's what initially confused me click ...that they used y1 as it is.

But I think I understand now. It's just another way of writing an equation of a line. In the first example it is written in a slope-intercept form, which is y = mx + b. And thus here we must first find the y-intercept m and then plug it back into the original equation.

And in the second example it is written in a point-slope formula y = m(x − x1) + y1. Because I guess it is more convenient in this case, since m here would equal −2168140,17647058... But the equation is still valid, just not very practical.

So I figure I could write the equation from the initial example as: y = −0.8(x − 1.5) − 2, which gives the same y = −0.8x −0.8 (oh wait, so then there was no need to plug 0,b trying to find m)...



Is it correct?

Check whether it gives the right answer for two different values of x. If it does, it must be right.