Equation of Line in Second Quadrant with Area 4 and Differing Intercepts

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Discussion Overview

The discussion revolves around finding the equation of a line that forms a triangle of area 4 with the axes in the second quadrant, while also having intercepts that differ by 5. The focus is on the mathematical formulation and derivation of the conditions necessary to solve the problem.

Discussion Character

  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents the two-intercept form of a line and sets up equations based on the area of the triangle and the difference in intercepts.
  • Another participant derives that the product of the intercepts is -8 and substitutes this into the equation relating the intercepts, leading to a quadratic equation.
  • The quadratic equation is shown to have a negative discriminant, indicating that there are no real solutions to the problem as formulated.

Areas of Agreement / Disagreement

Participants appear to agree on the mathematical steps taken, but there is no consensus on the existence of a solution, as the final conclusion indicates no real solutions are found.

Contextual Notes

The discussion highlights the dependence on the assumptions made regarding the intercepts and the area, as well as the implications of the negative discriminant in the quadratic equation.

Joe_1234
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Find the equation of the line which forms with the axes in the second quadrant a triangle of area 4 and whose intercepts differ by 5.
 
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Joe_1234 said:
Find the equation of the line which forms with the axes in the second quadrant a triangle of area 4 and whose intercepts differ by 5.

I would begin with the two-intercept equation of a line:

$$\frac{x}{a}+\frac{y}{b}=1$$

Where:

$$(a-b)^2=5^2$$

$$\frac{1}{2}(-a)b=4$$

We have two equations in two unknowns...can you proceed (observing that \(a<0<b\))?
 
Let's follow up...we have:

$$ab=-8\implies b=-\frac{8}{a}$$

And so:

$$\left(a+\frac{8}{a}\right)^2=5^2$$

$$\frac{a^2+8}{a}=\pm5$$

$$a^2\pm5a+8=0$$

$$a=\frac{\pm5\pm\sqrt{5^2-32}}{2}$$

And since the discriminant is negative, we find there is no real solution.
 
MarkFL said:
Let's follow up...we have:

$$ab=-8\implies b=-\frac{8}{a}$$

And so:

$$\left(a+\frac{8}{a}\right)^2=5^2$$

$$\frac{a^2+8}{a}=\pm5$$

$$a^2\pm5a+8=0$$

$$a=\frac{\pm5\pm\sqrt{5^2-32}}{2}$$

And since the discriminant is negative, we find there is no real solution.
Thank you sir
 

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