Soru

Zorluk: Çok zorLinear Functions and Graphs

In the xyxy-plane, line LL passes through the points (0,4)(0, 4) and (6,0)(6, 0). A second line, MM, is perpendicular to line LL and intersects line LL at a point on the line y=xy = x. If line MM is represented by the equation y=px+qy = px + q, where pp and qq are constants, what is the value of p+qp + q?

  1. A
    65\frac{6}{5}
  2. B
    103\frac{10}{3}
  3. 310\frac{3}{10}Cevap
  4. D
    −92-\frac{9}{2}

Cevap

The correct answer is the value stating three-tenths
To find the sum of the slope pp and yy-intercept qq of line MM, we first determine the equation of line LL. Line LL passes through (0,4)(0, 4) and (6,0)(6, 0), so its slope is 0−46−0=−23\frac{0 - 4}{6 - 0} = -\frac{2}{3}. The equation of line LL is therefore y=−23x+4y = -\frac{2}{3}x + 4. Setting y=xy = x gives x=−23x+4x = -\frac{2}{3}x + 4, which yields x=2.4x = 2.4. The intersection point is (2.4,2.4)(2.4, 2.4). Because line MM is perpendicular to line LL, its slope is the negative reciprocal of −23-\frac{2}{3}, which is p=1.5p = 1.5. Using point-slope form with the intersection point (2.4,2.4)(2.4, 2.4), the equation of line MM is y−2.4=1.5(x−2.4)y - 2.4 = 1.5(x - 2.4), which simplifies to y=1.5x−1.2y = 1.5x - 1.2. Thus, q=−1.2q = -1.2. The sum p+qp + q is 1.5−1.2=0.31.5 - 1.2 = 0.3, or 310\frac{3}{10}.

Adım Adım Çözüm

1
Find the equation of line LL using the given points (0,4)(0, 4) and (6,0)(6, 0).
The slope of line LL is mL=0−46−0=−23m_L = \frac{0 - 4}{6 - 0} = -\frac{2}{3}. Since the yy-intercept is (0,4)(0, 4), the equation of line LL is y=−23x+4y = -\frac{2}{3}x + 4.
To find where line LL intersects another line, we first need to establish its linear equation.
2
Find the intersection point of line LL and the line y=xy = x.
Substitute y=xy = x into the equation for line LL: x=−23x+4  ⟹  53x=4  ⟹  x=2.4x = -\frac{2}{3}x + 4 \implies \frac{5}{3}x = 4 \implies x = 2.4. Thus, the intersection point is (2.4,2.4)(2.4, 2.4).
The problem states that line MM intersects line LL at a point on the line y=xy = x.
3
Determine the equation of line MM using the intersection point and the perpendicular relationship.
Since line MM is perpendicular to line LL, its slope pp is the negative reciprocal of −23-\frac{2}{3}, which is p=32=1.5p = \frac{3}{2} = 1.5. Using point-slope form at (2.4,2.4)(2.4, 2.4): y−2.4=1.5(x−2.4)  ⟹  y=1.5x−1.2y - 2.4 = 1.5(x - 2.4) \implies y = 1.5x - 1.2. This gives q=−1.2q = -1.2.
Perpendicular lines in the coordinate plane have slopes that multiply to negative one. We then use the point-slope formula to define the equation of line MM.
4
Calculate the sum of pp and qq.
p+q=1.5+(−1.2)=0.3=310p + q = 1.5 + (-1.2) = 0.3 = \frac{3}{10}.
The question asks for the value of the expression p+qp + q.

Anahtar Kavram

Perpendicular lines and linear graph intersections
Tahmini Süre:3m 0s
Bu soruyu puanla