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Zorluk: ZorCoordinate Geometry and Lines

In the xyxy-plane, line LL passes through the points (t,t+2)(t, t+2) and (3t1,2t+7)(3t-1, 2t+7), where tt is a constant. Line MM is perpendicular to line LL and passes through the point (4,1)(4, -1). If the yy-intercept of line MM is 1111, what is the value of tt?

  1. 16-16Cevap
  2. B
    145-\frac{14}{5}
  3. C
    85\frac{8}{5}
  4. D
    27-\frac{2}{7}
  5. E
    1414

Cevap

16-16
The slope of line M is computed from its yy-intercept (0,11)(0, 11) and the point (4,1)(4, -1) as 11(1)04=3\frac{11 - (-1)}{0 - 4} = -3. Because line L is perpendicular to line M, the slope of line L is the negative reciprocal of 3-3, which is 13\frac{1}{3}. Calculating the slope of line L using points (t,t+2)(t, t+2) and (3t1,2t+7)(3t-1, 2t+7) gives (2t+7)(t+2)(3t1)t=t+52t1\frac{(2t+7)-(t+2)}{(3t-1)-t} = \frac{t+5}{2t-1}. Setting t+52t1=13\frac{t+5}{2t-1} = \frac{1}{3} leads to 3t+15=2t13t + 15 = 2t - 1, which simplifies to t=16t = -16.

Adım Adım Çözüm

1
Determine the slope of line M using its given points.
Line M passes through (4,1)(4, -1) and its yy-intercept (0,11)(0, 11). The slope mM=11(1)04=124=3m_M = \frac{11 - (-1)}{0 - 4} = \frac{12}{-4} = -3.
The slope of a line through (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by y2y1x2x1\frac{y_2 - y_1}{x_2 - x_1}.
2
Determine the required slope of line L.
Since line L is perpendicular to line M, its slope mL=1mM=13=13m_L = -\frac{1}{m_M} = -\frac{1}{-3} = \frac{1}{3}.
Perpendicular lines in the coordinate plane have slopes that are negative reciprocals of each other.
3
Express the slope of line L in terms of tt and solve for tt.
Using points (t,t+2)(t, t+2) and (3t1,2t+7)(3t-1, 2t+7), mL=(2t+7)(t+2)(3t1)t=t+52t1m_L = \frac{(2t+7) - (t+2)}{(3t-1) - t} = \frac{t+5}{2t-1}. Setting t+52t1=13\frac{t+5}{2t-1} = \frac{1}{3} yields 3(t+5)=1(2t1)    3t+15=2t1    t=163(t+5) = 1(2t-1) \implies 3t + 15 = 2t - 1 \implies t = -16.
Equating the algebraic slope expression to the numerical slope allows solving for the unknown parameter tt.

Anahtar Kavram

Perpendicular Slopes and Coordinate Line Equations
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