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Difficulty: HardLinear Equations in Two Variables

In the xyxy-plane, a line has an xx-intercept of (k+2,0)(k + 2, 0) and a yy-intercept of (0,2k+2)(0, 2k + 2), where kk is a positive constant. If the line passes through the point (3,5)(3, 5), what is the value of kk?

  1. A
    1.5
  2. B
    2.5
  3. 4Answer
  4. D
    6.5

Answer

4
The correct answer is the value 4. By substituting the point (3,5)(3, 5) into the intercept form of the line equation xk+2+y2k+2=1\frac{x}{k+2} + \frac{y}{2k+2} = 1, we obtain 3k+2+52k+2=1\frac{3}{k+2} + \frac{5}{2k+2} = 1. Multiplying by the common denominator (k+2)(2k+2)(k+2)(2k+2) and simplifying results in the quadratic equation 2k2−5k−12=02k^2 - 5k - 12 = 0. Factoring this equation yields (2k+3)(k−4)=0(2k+3)(k-4) = 0. Since kk must be a positive constant, we select k=4k = 4.

Step-by-Step Solution

1
Write the equation of the line using intercept form.
The equation of a line with xx-intercept (k+2,0)(k+2, 0) and yy-intercept (0,2k+2)(0, 2k+2) is xk+2+y2k+2=1\frac{x}{k+2} + \frac{y}{2k+2} = 1.
Intercept form xa+yb=1\frac{x}{a} + \frac{y}{b} = 1 is the most direct way to represent a line when both intercepts are given.
2
Substitute the given point (3,5)(3, 5) into the line's equation.
Substituting x=3x = 3 and y=5y = 5 yields the equation 3k+2+52k+2=1\frac{3}{k+2} + \frac{5}{2k+2} = 1.
Since the line passes through (3,5)(3, 5), these coordinates must satisfy the equation of the line.
3
Eliminate the denominators by multiplying both sides of the equation by the least common denominator.
Multiplying by (k+2)(2k+2)(k+2)(2k+2) gives 3(2k+2)+5(k+2)=(k+2)(2k+2)3(2k+2) + 5(k+2) = (k+2)(2k+2).
This clears the fractions and allows us to rewrite the equation in polynomial form.
4
Expand both sides of the equation and combine like terms.
Expanding both sides gives 6k+6+5k+10=2k2+2k+4k+4  ⟹  11k+16=2k2+6k+46k + 6 + 5k + 10 = 2k^2 + 2k + 4k + 4 \implies 11k + 16 = 2k^2 + 6k + 4.
To solve a polynomial equation, we need to simplify the algebraic expressions on both sides.
5
Rearrange the equation into standard quadratic form.
Moving all terms to one side gives 2k2−5k−12=02k^2 - 5k - 12 = 0.
Standard quadratic form Ax2+Bx+C=0Ax^2 + Bx + C = 0 is required to apply factoring or the quadratic formula.
6
Factor and solve the quadratic equation.
(2k+3)(k−4)=0(2k+3)(k-4) = 0, which gives k=−1.5k = -1.5 or k=4k = 4.
Factoring determines the values of kk that satisfy the quadratic relation.
7
Apply the constraint that kk is a positive constant.
Since k>0k > 0, we discard k=−1.5k = -1.5, leaving k=4k = 4.
The problem specifies that kk must be a positive constant.

Key Concept

Using the intercepts of a line to formulate its equation and solving the resulting rational/quadratic equation given a point on the line.

Alternative Method

Instead of using the intercept form of the line, you can equate the slope calculated between the y-intercept (0,2k+2)(0, 2k+2) and the point (3,5)(3,5) to the slope calculated between the point (3,5)(3,5) and the x-intercept (k+2,0)(k+2,0). This yields the equation 5−(2k+2)3−0=0−5(k+2)−3\frac{5 - (2k+2)}{3 - 0} = \frac{0 - 5}{(k+2) - 3}, which simplifies to 3−2k3=−5k−1\frac{3 - 2k}{3} = \frac{-5}{k - 1}. Cross-multiplying and simplifying leads to the same quadratic equation, 2k2−5k−12=02k^2 - 5k - 12 = 0.
Estimated Time:2m 30s
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Topics

Linear Equations in Two Variables

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