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Zorluk: ZorLinear Functions and Graphs

In the xyxy-plane, line ll passes through the point (3,4)(3, 4) and has a slope of mm, where m>0m > 0. Line kk is perpendicular to line ll and passes through the point (2,2)(2, 2). If the sum of the yy-intercepts of line ll and line kk is 11, what is the value of mm?

  1. A
    13\frac{1}{3}
  2. B
    23\frac{2}{3}
  3. C
    11
  4. 22Cevap

Cevap

The correct value of mm is 22.
The correct answer is the value 22. By finding the equations of both lines in slope-intercept form, we express their yy-intercepts in terms of mm: the yy-intercept of line ll is 43m4 - 3m and the yy-intercept of line kk is 2m+2\frac{2}{m} + 2. Setting their sum to 11 gives the equation 3m25m2=03m^2 - 5m - 2 = 0. Factoring this quadratic yields the solutions m=13m = -\frac{1}{3} and m=2m = 2. Since mm must be positive, m=2m = 2 is the only valid solution.

Adım Adım Çözüm

1
Find the equation and yy-intercept of line ll.
Using the point-slope form yy1=m(xx1)y - y_1 = m(x - x_1) with the point (3,4)(3, 4) and slope mm, the equation of line ll is y4=m(x3)y - 4 = m(x - 3), which simplifies to y=mx+43my = mx + 4 - 3m. Thus, the yy-intercept of line ll is 43m4 - 3m.
Expressing the yy-intercept of line ll in terms of mm allows us to use it in the sum equation.
2
Find the equation and yy-intercept of line kk.
Since line kk is perpendicular to line ll, its slope is the negative reciprocal of mm, which is 1m-\frac{1}{m}. Using the point-slope form with the point (2,2)(2, 2), the equation of line kk is y2=1m(x2)y - 2 = -\frac{1}{m}(x - 2), which simplifies to y=1mx+2m+2y = -\frac{1}{m}x + \frac{2}{m} + 2. Thus, the yy-intercept of line kk is 2m+2\frac{2}{m} + 2.
Expressing the yy-intercept of line kk in terms of mm allows us to use it in the sum equation.
3
Set up the equation for the sum of the yy-intercepts and solve for mm.
The sum of the yy-intercepts is (43m)+(2m+2)=1(4 - 3m) + (\frac{2}{m} + 2) = 1. Simplifying this equation gives 63m+2m=1    53m+2m=06 - 3m + \frac{2}{m} = 1 \implies 5 - 3m + \frac{2}{m} = 0. Multiplying by mm yields 3m25m2=03m^2 - 5m - 2 = 0. Factoring the quadratic gives (3m+1)(m2)=0(3m + 1)(m - 2) = 0, which has solutions m=13m = -\frac{1}{3} and m=2m = 2. Since m>0m > 0, we have m=2m = 2.
To determine the unique positive slope that satisfies the given conditions.

Anahtar Kavram

Writing equations of perpendicular lines and finding their intercepts using slopes and points.
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