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Zorluk: OrtaCoordinate Geometry: Lines, Slopes, and Distance

In the xyxy-plane, point M(3,3)M(3, 3) is the midpoint of line segment ABAB, where point AA has coordinates (1,2)(1, 2). Line LL is perpendicular to line segment ABAB and passes through point BB. What is the xx-intercept of line LL?

Cevap: 7

Cevap

7
Using the midpoint formula with A(1,2)A(1, 2) and M(3,3)M(3, 3) gives endpoint B(5,4)B(5, 4). The slope of segment ABAB is 4251=12\frac{4-2}{5-1} = \frac{1}{2}, so line LL, being perpendicular to ABAB, has slope 2-2. The equation of line LL passing through (5,4)(5, 4) is y4=2(x5)y - 4 = -2(x - 5), which simplifies to y=2x+14y = -2x + 14. Setting y=0y = 0 gives 0=2x+140 = -2x + 14, so the xx-intercept is 77.

Adım Adım Çözüm

1
Calculate the coordinates of endpoint B using the midpoint formula.
Point B has coordinates (5,4)(5, 4).
Midpoint M(xm,ym)=(xA+xB2,yA+yB2)M(x_m, y_m) = \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right). Solving 1+xB2=3\frac{1 + x_B}{2} = 3 yields xB=5x_B = 5, and solving 2+yB2=3\frac{2 + y_B}{2} = 3 yields yB=4y_B = 4.
2
Determine the slope of segment AB.
The slope mAB=12m_{AB} = \frac{1}{2}.
Using the slope formula m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1} between (1,2)(1, 2) and (5,4)(5, 4) gives 4251=24=12\frac{4 - 2}{5 - 1} = \frac{2}{4} = \frac{1}{2}.
3
Determine the slope of line L.
The slope of line L is 2-2.
Perpendicular lines have negative reciprocal slopes. The negative reciprocal of 12\frac{1}{2} is 2-2.
4
Find the equation of line L and solve for its x-intercept.
The x-intercept is 7.
Using point-slope form with B(5,4)B(5, 4) and m=2m = -2: y4=2(x5)    y=2x+14y - 4 = -2(x - 5) \implies y = -2x + 14. Setting y=0y = 0 gives 0=2x+14    x=70 = -2x + 14 \implies x = 7.

Anahtar Kavram

Midpoint Formula, Perpendicular Slopes, and Line Intercepts
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