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Difficulty: HardCoordinate Geometry: Lines, Slopes, and Distance

In the xyxy-plane, line kk passes through the origin (0,0)(0,0) and the point (4,3)(4, 3). Line mm is perpendicular to line kk and intersects line kk at point (4,3)(4, 3). If line mm intersects the xx-axis at point QQ and the yy-axis at point RR, what is the distance between point QQ and point RR?

  1. A
    3512\frac{35}{12}
  2. B
    2512\frac{25}{12}
  3. C
    1010
  4. 12512\frac{125}{12}Answer
  5. E
    17512\frac{175}{12}

Answer

12512\frac{125}{12}
The correct answer is derived by finding the slope of line kk as 34\frac{3}{4}, taking its negative reciprocal −43-\frac{4}{3} as the slope of perpendicular line mm, using the point (4,3)(4,3) to form line mm's equation y=−43x+253y = -\frac{4}{3}x + \frac{25}{3}, identifying intercepts Q(254,0)Q\left(\frac{25}{4}, 0\right) and R(0,253)R\left(0, \frac{25}{3}\right), and applying the distance formula to obtain (254)2+(253)2=12512\sqrt{\left(\frac{25}{4}\right)^2 + \left(\frac{25}{3}\right)^2} = \frac{125}{12}.

Step-by-Step Solution

1
Determine the slope of line kk
The slope of line kk is mk=3−04−0=34m_k = \frac{3 - 0}{4 - 0} = \frac{3}{4}.
The slope of a line passing through (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by y2−y1x2−x1\frac{y_2 - y_1}{x_2 - x_1}.
2
Find the equation of line mm
Line mm has slope mm=−43m_m = -\frac{4}{3} and equation y−3=−43(x−4)  ⟹  y=−43x+253y - 3 = -\frac{4}{3}(x - 4) \implies y = -\frac{4}{3}x + \frac{25}{3}.
Perpendicular lines have slopes that are negative reciprocals of each other (mm=−1mkm_m = -\frac{1}{m_k}). Point-slope form is used with point (4,3)(4, 3).
3
Calculate the intercepts QQ and RR
xx-intercept Q=(254,0)Q = \left(\frac{25}{4}, 0\right) and yy-intercept R=(0,253)R = \left(0, \frac{25}{3}\right).
Setting y=0y = 0 in y=−43x+253y = -\frac{4}{3}x + \frac{25}{3} gives x=254x = \frac{25}{4}. Setting x=0x = 0 gives y=253y = \frac{25}{3}.
4
Compute the distance between QQ and RR
Distance d=(254)2+(253)2=25116+19=2525144=25×512=12512d = \sqrt{\left(\frac{25}{4}\right)^2 + \left(\frac{25}{3}\right)^2} = 25 \sqrt{\frac{1}{16} + \frac{1}{9}} = 25 \sqrt{\frac{25}{144}} = 25 \times \frac{5}{12} = \frac{125}{12}.
Applying the distance formula d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} for points QQ and RR.

Key Concept

Perpendicular slopes and distance formula in coordinate geometry
Estimated Time:2m 0s
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Topics

Coordinate Geometry: Lines, Slopes, and Distance

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