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Zorluk: ZorLinear Equations and Graphing

In the standard (x,y)(x, y) coordinate plane, a line L1L_1 passes through the points (2,5)(2, 5) and (1,4)(-1, -4). A second line, L2L_2, passes through the yy-intercept of L1L_1 and has a slope that is twice the slope of L1L_1. What is the xx-intercept of L2L_2?

  1. 16\frac{1}{6}Cevap
  2. B
    16-\frac{1}{6}
  3. C
    13\frac{1}{3}
  4. D
    32-\frac{3}{2}
  5. E
    132-\frac{13}{2}

Cevap

The xx-intercept of L2L_2 is 16\frac{1}{6}.
To find the xx-intercept of L2L_2, the slope and yy-intercept of L1L_1 must first be found. The slope of L1L_1 is 5(4)2(1)=3\frac{5 - (-4)}{2 - (-1)} = 3. Using the point-slope form with the point (2,5)(2, 5), the equation of L1L_1 is y5=3(x2)    y=3x1y - 5 = 3(x - 2) \implies y = 3x - 1, meaning its yy-intercept is (0,1)(0, -1). The second line, L2L_2, has a slope of 2×3=62 \times 3 = 6 and passes through the same yy-intercept (0,1)(0, -1), giving the equation y=6x1y = 6x - 1. Setting y=0y = 0 to find the xx-intercept yields 6x1=0    x=166x - 1 = 0 \implies x = \frac{1}{6}.

Adım Adım Çözüm

1
Calculate the slope of line L1L_1 using the coordinates (2,5)(2, 5) and (1,4)(-1, -4).
The slope of L1L_1 is m1=5(4)2(1)=93=3m_1 = \frac{5 - (-4)}{2 - (-1)} = \frac{9}{3} = 3.
The slope of a line containing (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is defined as y2y1x2x1\frac{y_2 - y_1}{x_2 - x_1}.
2
Determine the equation and the yy-intercept of line L1L_1.
The equation of L1L_1 in point-slope form is y5=3(x2)    y=3x1y - 5 = 3(x - 2) \implies y = 3x - 1. The yy-intercept of L1L_1 is (0,1)(0, -1).
Writing the equation in slope-intercept form y=mx+by = mx + b directly identifies the yy-intercept at (0,b)(0, b).
3
Determine the equation of line L2L_2.
The slope of L2L_2 is 2×3=62 \times 3 = 6. Since L2L_2 passes through the yy-intercept of L1L_1 at (0,1)(0, -1), its equation is y=6x1y = 6x - 1.
Using the relationship for the doubled slope and the shared yy-intercept, the slope-intercept form of the second line is determined.
4
Find the xx-intercept of L2L_2 by setting y=0y = 0.
Setting y=0y = 0 in the equation y=6x1y = 6x - 1 gives 0=6x1    6x=1    x=160 = 6x - 1 \implies 6x = 1 \implies x = \frac{1}{6}.
The xx-intercept occurs where the line crosses the xx-axis, which corresponds to setting y=0y = 0.

Anahtar Kavram

Determining linear equations from coordinates, finding slopes, and calculating coordinate intercepts.
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