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Zorluk: ZorLinear Equations and Graphing

A line in the standard (x,y)(x,y) coordinate plane passes through the point (4,1)(4, 1) and has a negative slope mm. The line and the coordinate axes bound a region in the first quadrant. If the area of this region is 88 square units, which of the following is the value of mm?

  1. A
    −4-4
  2. B
    −916-\frac{9}{16}
  3. −14-\frac{1}{4}Cevap
  4. D
    −34-\frac{3}{4}
  5. E
    −98-\frac{9}{8}

Cevap

−14-\frac{1}{4}
The correct answer is −14-\frac{1}{4}. The line passes through (4,1)(4, 1) with slope mm. Using the point-slope formula, the equation of the line is y−1=m(x−4)y - 1 = m(x - 4), which simplifies to y=mx−4m+1y = mx - 4m + 1. The yy-intercept is found by setting x=0x = 0, giving 1−4m1 - 4m. The xx-intercept is found by setting y=0y = 0, giving 4m−1m\frac{4m-1}{m}. The area of the right-triangular region in the first quadrant bounded by the axes is 12⋅(base)⋅(height)=8\frac{1}{2} \cdot (\text{base}) \cdot (\text{height}) = 8. Substituting the intercepts, we get 12⋅(4m−1m)⋅(1−4m)=8\frac{1}{2} \cdot \left(\frac{4m-1}{m}\right) \cdot (1-4m) = 8. Multiplying by 2m2m (which is negative, so we maintain positive side lengths) yields −(1−4m)2=16m-(1-4m)^2 = 16m, which simplifies to 16m2+8m+1=016m^2 + 8m + 1 = 0. Factoring the quadratic expression gives (4m+1)2=0(4m+1)^2 = 0, which has the single solution m=−14m = -\frac{1}{4}.

Adım Adım Çözüm

1
Write the equation of the line using the point-slope form with the point (4,1)(4, 1) and slope mm.
y−1=m(x−4)  ⟹  y=mx−4m+1y - 1 = m(x - 4) \implies y = mx - 4m + 1
The point-slope formula y−y1=m(x−x1)y - y_1 = m(x - x_1) defines any line passing through a given point with a specific slope.
2
Find the xx-intercept and yy-intercept of the line.
y-intercept=1−4my\text{-intercept} = 1 - 4m (when x=0x=0), and x-intercept=4m−1mx\text{-intercept} = \frac{4m-1}{m} (when y=0y=0)
The intercepts represent the vertices of the right triangle formed by the line and the coordinate axes.
3
Set up the area equation for the right triangle in the first quadrant, noting that since m<0m < 0, both intercepts are positive.
Area=12⋅base⋅height  ⟹  12⋅(4m−1m)⋅(1−4m)=8Area = \frac{1}{2} \cdot \text{base} \cdot \text{height} \implies \frac{1}{2} \cdot \left(\frac{4m-1}{m}\right) \cdot (1-4m) = 8
The area of the region bounded by the axes and the line is a right triangle whose legs are the intercepts.
4
Solve the algebraic equation for mm.
(4m−1)(1−4m)=16m  ⟹  16m2+8m+1=0  ⟹  (4m+1)2=0  ⟹  m=−14(4m-1)(1-4m) = 16m \implies 16m^2 + 8m + 1 = 0 \implies (4m+1)^2 = 0 \implies m = -\frac{1}{4}
Multiplying both sides by 2m2m and expanding the terms leads to a quadratic equation in terms of mm, which resolves to a single real root.

Anahtar Kavram

Using linear equation forms and coordinate geometry to represent boundary lines and compute bounded areas.

Alternatif Yöntem

Instead of using the point-slope form, write the line in intercept form: xa+yb=1\frac{x}{a} + \frac{y}{b} = 1, where aa and bb are the xx- and yy-intercepts. The area of the region is 12ab=8  ⟹  ab=16  ⟹  b=16a\frac{1}{2}ab = 8 \implies ab = 16 \implies b = \frac{16}{a}. Substitute this back into the intercept form to get xa+y16/a=1  ⟹  xa+ay16=1\frac{x}{a} + \frac{y}{16/a} = 1 \implies \frac{x}{a} + \frac{ay}{16} = 1. Since the line passes through (4,1)(4, 1), substitute these coordinates: 4a+a(1)16=1\frac{4}{a} + \frac{a(1)}{16} = 1. Multiply the entire equation by 16a16a to clear the denominators: 64+a2=16a  ⟹  a2−16a+64=0  ⟹  (a−8)2=0  ⟹  a=864 + a^2 = 16a \implies a^2 - 16a + 64 = 0 \implies (a-8)^2 = 0 \implies a = 8. Since a=8a = 8, the yy-intercept is b=168=2b = \frac{16}{8} = 2. Using the intercepts (8,0)(8, 0) and (0,2)(0, 2), the slope is m=2−00−8=−28=−14m = \frac{2 - 0}{0 - 8} = -\frac{2}{8} = -\frac{1}{4}.
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Linear Equations and Graphing

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