Math  /  Algebra

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Question What is the equation of the line that passes through the point (6,3)(-6,-3) and has a slope of 16-\frac{1}{6} ?
Answer Attempt 1 out of 2

Studdy Solution

STEP 1

What is this asking? We need to find the equation of a line given a **point** it passes through and its **slope**. Watch out! Don't mix up the xx and yy coordinates of the point!

STEP 2

1. Point-Slope Form
2. Simplify to Slope-Intercept Form

STEP 3

Alright, let's **start** with the point-slope form of a linear equation, which is perfect for when we know a point (x1,y1)(x_1, y_1) on the line and the slope mm of the line.
The formula is: yy1=m(xx1)y - y_1 = m(x - x_1) This formula is awesome because it tells us how much the yy-value changes based on how far we move the xx-value.

STEP 4

We know our point is (6,3)(-6, -3), so x1=6x_1 = -6 and y1=3y_1 = -3.
Our slope is m=16m = -\frac{1}{6}.
Let's **plug these values** into the point-slope form: y(3)=16(x(6))y - (-3) = -\frac{1}{6}(x - (-6))

STEP 5

Let's **clean this up** a bit!
We can simplify the equation to get it into slope-intercept form (y=mx+by = mx + b), which is generally more useful.
First, those double negatives become positives: y+3=16(x+6)y + 3 = -\frac{1}{6}(x + 6)

STEP 6

Now, **distribute** that 16-\frac{1}{6} to both terms inside the parentheses: y+3=16x+(166)y + 3 = -\frac{1}{6} \cdot x + \left( -\frac{1}{6} \cdot 6 \right) y+3=16x1y + 3 = -\frac{1}{6}x - 1

STEP 7

Almost there!
We want to **isolate** yy, so let's subtract 3 from both sides of the equation: y+33=16x13y + 3 - 3 = -\frac{1}{6}x - 1 - 3 y=16x4y = -\frac{1}{6}x - 4

STEP 8

The equation of the line is y=16x4y = -\frac{1}{6}x - 4.
Boom!

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