Math Problem: Is Devon's Line Equation A Good Model?

Dec 20, 2025 by Andrew McMorgan 53 views

Hey guys, welcome back to Plastik Magazine! Today, we've got a cool math problem that involves figuring out if a given equation actually fits some collected data points. It's all about line equations and checking if a model is a good fit. So, let's dive into it!

The Challenge: Evaluating a Line Equation Model

Devon's collected some data and he's trying to model it with a straight line. He's got two specific data points that his line should pass through: $(8,5)$ and $(-12,-9)$ . Based on these points, he's come up with the equation $7x - 10y = 3$ . Now, the big question is: is this a good model? We need to break this down and explain our reasoning, so everyone can understand why or why not.

Understanding Line Equations and Data Points

Alright, let's talk about what it means for a line to pass through data points. When we have a line defined by an equation, like $Ax + By = C$ or $y = mx + b$ , any point $(x,y)$ that lies on that line will satisfy the equation. This means if you plug the x and y coordinates of a point into the equation, the equation will hold true. For Devon's model to be a good model based on the two points he mentioned, both of those points must satisfy his equation $7x - 10y = 3$ . If even one of those points doesn't work, then his equation isn't accurately representing the line that passes through those specific data points.

To check this, we're going to perform a simple substitution. We'll take the coordinates of the first point, $(8,5)$ , and plug them into Devon's equation. Then, we'll do the same for the second point, $(-12,-9)$ . If both substitutions result in a true statement (i.e., the left side of the equation equals the right side), then Devon's equation is indeed a good model for a line passing through those two points. If either substitution leads to a false statement, then his model is not good, and we'll need to figure out why.

The first point we need to test is $(8,5)$ . This means $x=8$ and $y=5$ . Let's substitute these values into Devon's equation: $7x - 10y = 3$ .

So, we get: $7(8) - 10(5)$ .

Calculating this, we have $56 - 50$ .

This simplifies to $6$ .

Now, we compare this result to the right side of Devon's equation, which is $3$ . So, the equation becomes $6 = 3$ . Is this true? Nope! $6$ does not equal $3$ . This immediately tells us that the point $(8,5)$ does not lie on the line represented by the equation $7x - 10y = 3$ . This alone is enough to say that Devon's model is not good for a line passing through these two specific points.

However, to be thorough and to really understand the situation, let's check the second point as well. The second point is $(-12,-9)$ . Here, $x=-12$ and $y=-9$ . Let's substitute these into $7x - 10y = 3$ :

$7(-12) - 10(-9)$ .

Multiplying these out, we get $-84 - (-90)$ .

Subtracting a negative is the same as adding a positive, so this becomes $-84 + 90$ .

And $-84 + 90$ equals $6$ .

Again, we compare this result to the right side of Devon's equation, which is $3$ . So, the equation becomes $6 = 3$ . Is this true? Nope again! Just like the first point, the second point $(-12,-9)$ also does not lie on the line represented by the equation $7x - 10y = 3$ .

Since neither of the data points Devon collected satisfies his equation, his model $7x - 10y = 3$ is definitely not a good model for a line that passes through the points $(8,5)$ and $(-12,-9)$ . The reasoning is straightforward: a line equation is only valid for points that lie on that line, and our tests show these points don't. They don't even satisfy it partially; both substitutions resulted in $6$ , which is not equal to $3$ . This consistency in the wrong answer is interesting, and we'll explore that a bit more in a moment.

Finding the Correct Line Equation

So, Devon's equation didn't work out. That's okay! Math is all about trying, checking, and refining. Now, let's figure out what the actual equation of the line passing through $(8,5)$ and $(-12,-9)$ should be. To do this, we first need to calculate the slope of the line. The formula for slope ( $m$ ) between two points $(x_1, y_1)$ and $(x_2, y_2)$ is: $m = rac{y_2 - y_1}{x_2 - x_1}$ .

Let's use $(x_1, y_1) = (8,5)$ and $(x_2, y_2) = (-12,-9)$ .

Plugging these into the slope formula, we get:

$m = rac{-9 - 5}{-12 - 8}$

$m = rac{-14}{-20}$

We can simplify this fraction by dividing both the numerator and the denominator by $-2$ . This gives us:

$m = rac{7}{10}$

So, the slope of the line passing through these two points is $rac{7}{10}$ . This means for every 10 units we move to the right on the x-axis, the line goes up 7 units on the y-axis. This is a crucial piece of information for finding the equation of the line.

Now that we have the slope, we can use the point-slope form of a linear equation, which is $y - y_1 = m(x - x_1)$ . We can use either of our two points for $(x_1, y_1)$ . Let's use the point $(8,5)$ for this calculation.

Substituting $m = rac{7}{10}$ and $(x_1, y_1) = (8,5)$ into the point-slope form, we get:

$y - 5 = rac{7}{10}(x - 8)$

Now, let's rearrange this equation to get it into the standard form ( $Ax + By = C$ ), similar to Devon's original equation. First, distribute the $rac{7}{10}$ on the right side:

$y - 5 = rac{7}{10}x - rac{7}{10}(8)$

$y - 5 = rac{7}{10}x - rac{56}{10}$

We can simplify $rac{56}{10}$ to $rac{28}{5}$ .

$y - 5 = rac{7}{10}x - rac{28}{5}$

To get rid of the fractions, we can multiply the entire equation by the least common denominator, which is 10:

$10(y - 5) = 10( rac{7}{10}x - rac{28}{5})$

$10y - 50 = 7x - 10( rac{28}{5})$

$10y - 50 = 7x - 2(28)$

$10y - 50 = 7x - 56$

Now, we want to rearrange this into the form $Ax + By = C$ . Let's move the $7x$ term to the left side and the $-50$ term to the right side:

$-7x + 10y = -56 + 50$

$-7x + 10y = -6$

This is a valid equation for the line. However, it's often preferred to have the coefficient of $x$ be positive. We can multiply the entire equation by $-1$ to achieve this:

$7x - 10y = 6$

So, the correct equation for the line passing through $(8,5)$ and $(-12,-9)$ is $7x - 10y = 6$ . Compare this to Devon's equation: $7x - 10y = 3$ . They have the same left side ( $7x - 10y$ ), but the right side is different (6 vs. 3).

Why Did Devon's Equation Give the Same Incorrect Result?

It's super interesting that when we plugged both points $(8,5)$ and $(-12,-9)$ into Devon's equation $7x - 10y = 3$ , we consistently got $6$ on the left side, leading to the false statement $6 = 3$ . This isn't just a random coincidence; it tells us something important about the relationship between Devon's equation and the actual line.

Remember when we calculated the slope of the correct line? We found it to be $m = rac{7}{10}$ . Now, let's look at the structure of Devon's equation, $7x - 10y = 3$ . If we rearrange this into the slope-intercept form ( $y = mx + b$ ), we can see its slope.

Starting with $7x - 10y = 3$ :

Subtract $7x$ from both sides:

$-10y = -7x + 3$

Divide by $-10$ :

$y = rac{-7x}{-10} + rac{3}{-10}$

$y = rac{7}{10}x - rac{3}{10}$

Look at that! The slope of Devon's line is also $rac{7}{10}$ . This is exactly the same slope as the line that actually passes through the two points! What this means is that Devon's equation represents a line that is parallel to the correct line. Parallel lines have the same slope but never intersect.

Devon's equation $7x - 10y = 3$ represents a line, and the points $(8,5)$ and $(-12,-9)$ lie on a different line that is parallel to it. When we plugged the points into $7x - 10y = 3$ , we were essentially calculating the value of $7x - 10y$ for those points. Since the structure $7x - 10y$ is preserved in both the correct line ( $7x - 10y = 6$ ) and Devon's line ( $7x - 10y = 3$ ), the points on the correct line will always result in the same value for $7x - 10y$ . For the points $(8,5)$ and $(-12,-9)$ , this value turned out to be $6$ . Devon's equation requires this value to be $3$ , which is why his model is incorrect. The constant term (the 'C' in $Ax + By = C$ ) determines where the line is positioned relative to the origin, and Devon's constant term is off.

Conclusion: Why Devon's Model Isn't Good

To wrap things up, guys, Devon's model $7x - 10y = 3$ is not a good model for a line passing through the data points $(8,5)$ and $(-12,-9)$ .

Our reasoning is based on two key checks:

Substitution Check: When we substituted the coordinates of the given points into Devon's equation, neither point satisfied the equation. Specifically, for $(8,5)$ , we got $6 = 3$ , and for $(-12,-9)$ , we also got $6 = 3$ . For a point to lie on a line, it must satisfy the line's equation.
Correct Equation Derivation: By calculating the slope and using the point-slope form, we derived the correct equation for the line passing through the given points, which is $7x - 10y = 6$ . This clearly shows that Devon's equation, with its constant term of 3 instead of 6, is incorrect.

Interestingly, Devon's equation represents a line parallel to the correct line, sharing the same slope ( $rac{7}{10}$ ). This explains why both points yielded the same incorrect result ( $6$ ) when tested against his equation. The value $7x - 10y$ for these points is $6$ , not $3$ .

So, while Devon was on the right track by recognizing the need for a linear equation and correctly identifying the slope structure, the final constant term made his model inaccurate for the specified data points. Keep practicing, keep checking your work, and you'll nail these problems every time!