How do I find an equation of the line with a given x-intercept and point?

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SUMMARY

The discussion focuses on finding the equation of a line that passes through the point (6, 2) and shares the same x-intercept as the line represented by the equation -2x + y = 1. The x-intercept is determined to be (-1/2, 0) by setting y to 0 in the original equation. Using the point-slope formula, the slope is calculated from the two points, and the final equation of the line is derived by substituting the slope and one of the points into the formula. The process emphasizes the importance of understanding the relationship between points and slopes in linear equations.

PREREQUISITES
  • Understanding of linear equations and their forms, specifically y = ax + b.
  • Familiarity with the point-slope formula: y - y1 = m(x - x1).
  • Ability to calculate slopes between two points.
  • Knowledge of finding x-intercepts by setting y to zero in an equation.
NEXT STEPS
  • Practice using the point-slope formula with different sets of points.
  • Explore the concept of slope-intercept form and its applications.
  • Learn how to derive equations from given points and slopes in various contexts.
  • Investigate the significance of x-intercepts and y-intercepts in graphing linear equations.
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Students learning algebra, educators teaching linear equations, and anyone interested in mastering the fundamentals of line equations in coordinate geometry.

mathdad
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Find an equation of the line that passes through (6, 2) and has the same x-intercept as the line -2x + y = 1.

As a first step, I must let y = 0 in the given equation.

-2x + y = 1

-2x + 0 = 1

-2x = 1

x = -1/2

The x-intercept is (-1/2, 0) creating the second point needed to find the slope.

What is next?
 
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Okay, you now have two points on the line, so you can compute the slope, and then you have the slope and point(s) on the line, so the point-slope formula will be useful:

$$y-y_1=m\left(x-x_1\right)$$

:D
 
MarkFL said:
Okay, you now have two points on the line, so you can compute the slope, and then you have the slope and point(s) on the line, so the point-slope formula will be useful:

$$y-y_1=m\left(x-x_1\right)$$

:D

Perfect. It's easier than I thought.
 
Equivalently, any (non-vertical) line can be written in the form y= ax+ b. Knowing that the line goes through (-1/2, 0) tells you that 0= a(-1/2)+ b. Knowing that the line goes through (6, 2) tells you that 2= a(6)+ b, giving two equations to solve for a and b. You can immediately eliminate b by subtracting one equation from the other which gives precisely the previous method.
 
HallsofIvy said:
Equivalently, any (non-vertical) line can be written in the form y= ax+ b. Knowing that the line goes through (-1/2, 0) tells you that 0= a(-1/2)+ b. Knowing that the line goes through (6, 2) tells you that 2= a(6)+ b, giving two equations to solve for a and b. You can immediately eliminate b by subtracting one equation from the other which gives precisely the previous method.
Good data here.
 
Find an equation of the line that passes through (6, 2) and has the same x-intercept as the line -2x + y = 1.

As a first step, I must let y = 0 in the given equation.

-2x + y = 1

-2x + 0 = 1

-2x = 1

x = -1/2

The x-intercept is (-1/2, 0) creating the second point needed to find the slope.

I will now solve the given equation for y.

-2x + y = 1

y= 2x + 1

The equation I need to find can be found by finding the slope using (-1/2, 0) and (6, 2).

I then plug one of the points and the slope into the point-slope formula. As a last step, solve the equation for y.
 

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