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Difficulty: HardLinear Equations and Graphing

In the standard (x,y)(x, y) coordinate plane, a line L1L_1 with a positive slope mm and a negative yy-intercept bb passes through the point (4,3)(4, 3). The region in the fourth quadrant bounded by the line L1L_1, the xx-axis, and the yy-axis has an area of exactly 88 square units. What is the yy-coordinate of the yy-intercept of line L1L_1?

Answer: -6

Answer

The y-coordinate of the y-intercept of line L1L_1 is −6-6.
The correct answer is −6-6. Substituting (4,3)(4, 3) into the slope-intercept equation y=mx+by = mx + b gives 3=4m+b3 = 4m + b, or m=3−b4m = \frac{3-b}{4}. The area of the right triangle in the fourth quadrant is 12×base×height=12(−bm)(−b)=b22m=8\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} (-\frac{b}{m})(-b) = \frac{b^2}{2m} = 8. Substituting mm yields b2=16(3−b4)b^2 = 16\left(\frac{3-b}{4}\right), which simplifies to the quadratic equation b2+4b−12=0b^2 + 4b - 12 = 0. Factoring gives (b+6)(b−2)=0(b+6)(b-2) = 0. Since the y-intercept bb must be negative, we have b=−6b = -6.

Step-by-Step Solution

1
Substitute the given point into the slope-intercept equation
m=3−b4m = \frac{3 - b}{4}
Since the line passes through (4,3)(4, 3), substituting these coordinates into y=mx+by = mx + b allows us to express the slope mm in terms of the y-intercept bb.
2
Determine the intercepts and the dimensions of the bounded region
Base =−bm= -\frac{b}{m} and Height =−b= -b
The boundary of the region in the fourth quadrant is a right triangle formed by the origin, the x-intercept (−bm,0)(-\frac{b}{m}, 0), and the y-intercept (0,b)(0, b).
3
Set up the area of the triangle and equate it to 8
b2=16mb^2 = 16m
The area of a right triangle is 12×base×height\frac{1}{2} \times \text{base} \times \text{height}, so 12(−bm)(−b)=8\frac{1}{2} \left(-\frac{b}{m}\right)(-b) = 8 simplifies to b2=16mb^2 = 16m.
4
Substitute mm into the area equation and solve the resulting quadratic equation
b=−6b = -6 (discarding b=2b = 2)
Substituting m=3−b4m = \frac{3 - b}{4} yields b2+4b−12=0b^2 + 4b - 12 = 0, which factors into (b+6)(b−2)=0(b + 6)(b - 2) = 0. Since the region is in the fourth quadrant, the y-intercept must be negative (b<0b < 0).

Key Concept

Using linear equation intercepts to calculate bounded areas on the coordinate plane and relating variables using point substitution.
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Topics

Linear Equations and Graphing

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