Find the equation of the line that is parallel to $y=-7 x+2$ and contains the point $(-5,32)$.
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Answer (2)
The line is parallel to y = − 7 x + 2 , so it has a slope of − 7 .
The equation is of the form y = − 7 x + b .
Substitute the point ( − 5 , 32 ) into the equation: 32 = − 7 ( − 5 ) + b .
Solve for b : b = − 3 , so the equation is y = − 7 x − 3 .
Explanation
Understanding the Problem We are given a line y = − 7 x + 2 and a point ( − 5 , 32 ) . We need to find the equation of a line that is parallel to the given line and passes through the given point.
Finding the Slope Since the line we are looking for is parallel to y = − 7 x + 2 , it will have the same slope. The slope of the given line is − 7 . Therefore, the equation of the line we are looking for will be of the form y = − 7 x + b , where b is the y-intercept.
Substituting the Point Now, we need to find the value of b . We know that the line passes through the point ( − 5 , 32 ) . We can substitute these coordinates into the equation y = − 7 x + b to solve for b . So, we have 32 = − 7 ( − 5 ) + b .
Solving for the y-intercept Let's simplify the equation: 32 = 35 + b . To solve for b , we subtract 35 from both sides of the equation: b = 32 − 35 = − 3 .
Writing the Equation Now that we have the slope − 7 and the y-intercept − 3 , we can write the equation of the line as y = − 7 x − 3 .
Examples
Imagine you're designing a ramp for a skateboard park. You want the ramp to have the same steepness as another ramp ( y = − 7 x + 2 ) but need it to start at a different point in the park ( − 5 , 32 ). By finding a parallel line that passes through the new starting point, you ensure the ramp has the desired steepness while fitting perfectly into the park's layout. This problem demonstrates how understanding parallel lines helps in design and construction, ensuring consistency and functionality.
The equation of the line parallel to y = − 7 x + 2 and passing through the point ( − 5 , 32 ) is y = − 7 x − 3 . The slope of − 7 is the same as the given line, and we found the y-intercept by substituting the coordinates of the point. The final equation captures the relationship desired.
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