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Slope of line based on one point and area of triangle

  1. Mar 8, 2009 #1
    1. The problem statement, all variables and given/known data
    The problem is shown as a picture, but here it is in word form: A straight line with a negative slope intersects the point 2,1. The area under this line in quadrant one of the cartesian grid is 4. What is the slope of this line?

    2. Relevant equations
    Area of a triangle: A=0.5bh
    Point-Slope formula: y-y1=m(x-x1)
    Slope: m=(y2-y1)/(x2-x1)

    3. The attempt at a solution
    I wrote a lot of equations that met the criterion of passing through the point 2,1 and having a negative slope, and calculated the area of the triangle the line created in quadrant one. I came to the solution this way - m=-0.5. However, I want to find a more elegant way of approaching this solution.
    Last edited: Mar 8, 2009
  2. jcsd
  3. Mar 8, 2009 #2


    Staff: Mentor

    Using the given point, an equation of the line is y - 1 = m(x - 2), or y = mx - 2m + 1.
    Use the equation of this line to find the x- and y-intercepts. These will be the base and altitude of your triangle. Since you don't know m (the slope of the line), both will be in terms of m

    After you have found the intercepts, write a new equation that represents the area of the triangle.

    4 = 1/2 * (x-intercept)*(y-intercept)

    The equation you get can be made into a quadratic equation, and its only solution is m = -1/2, which is in agreement with the value you already found.
  4. Mar 8, 2009 #3
    Thank you! That is exactly what I was looking for.
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