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Zorluk: ZorParallel and Perpendicular Lines

A line, L1L_1, is perpendicular to a second line whose equation is 5x2y=155x - 2y = 15. The line L1L_1 intersects the yy-axis at (0,12)(0, 12) and passes through the midpoint of a line segment with endpoints at (3,k)(3, k) and (7,6)(7, 6) in the standard (x,y)(x, y) coordinate plane. What is the value of kk?

Cevap: 14

Cevap

The correct value of kk is 14.
The slope of the line 5x2y=155x - 2y = 15 is found by solving for yy, yielding y=52x7.5y = \frac{5}{2}x - 7.5. The slope of any perpendicular line is the negative reciprocal of 52\frac{5}{2}, which is 25-\frac{2}{5}. Given the yy-intercept (0,12)(0, 12), the equation of the perpendicular line L1L_1 is y=25x+12y = -\frac{2}{5}x + 12. The midpoint of the segment with endpoints (3,k)(3, k) and (7,6)(7, 6) is calculated as (3+72,k+62)=(5,k+62)\left(\frac{3+7}{2}, \frac{k+6}{2}\right) = \left(5, \frac{k+6}{2}\right). Since the midpoint lies on L1L_1, substituting x=5x = 5 into the line equation gives y=25(5)+12=10y = -\frac{2}{5}(5) + 12 = 10. Equating this to the midpoint's yy-coordinate expression gives k+62=10\frac{k+6}{2} = 10, which solves to k=14k = 14.

Adım Adım Çözüm

1
Find the slope of the line 5x2y=155x - 2y = 15 by converting it to slope-intercept form (y=mx+by = mx + b).
The slope of the line is 52\frac{5}{2}.
Rewriting the equation as 2y=5x+15-2y = -5x + 15 and dividing by 2-2 isolates yy and reveals the slope.
2
Determine the perpendicular slope for line L1L_1.
The slope of L1L_1 is 25-\frac{2}{5}.
Perpendicular lines have slopes that are negative reciprocals of one another.
3
Formulate the equation of line L1L_1 using its slope and the given yy-intercept (0,12)(0, 12).
The equation of L1L_1 is y=25x+12y = -\frac{2}{5}x + 12.
The slope-intercept form is y=mx+by = mx + b, where mm is the slope and bb is the yy-coordinate of the yy-intercept.
4
Find the midpoint of the line segment with endpoints (3,k)(3, k) and (7,6)(7, 6) in terms of kk.
The midpoint is (5,k+62)\left(5, \frac{k+6}{2}\right).
The midpoint formula calculates the average of the xx-coordinates and the average of the yy-coordinates.
5
Substitute the midpoint coordinates into the equation of L1L_1 and solve for kk.
k=14k = 14
Since the midpoint lies on line L1L_1, substituting its xx and yy values into the equation must satisfy the equality.

Anahtar Kavram

Using perpendicular slopes and the midpoint formula to determine unknown coordinate values.
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