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Zorluk: ZorParallel and Perpendicular Lines

In the standard (x,y)(x, y) coordinate plane, a triangle has vertices at A(k,4)A(k, 4), B(−1,2)B(-1, 2), and C(3,−6)C(3, -6). The altitude from vertex AA to side BCBC intersects the yy-axis at (0,−2)(0, -2). What is the value of kk?

  1. A
    -12
  2. B
    3
  3. C
    4
  4. 12Cevap
  5. E
    24

Cevap

12
The slope of side BCBC is calculated as mBC=−6−23−(−1)=−2m_{BC} = \frac{-6 - 2}{3 - (-1)} = -2. Since the altitude from vertex AA is perpendicular to side BCBC, its slope must be the negative reciprocal of −2-2, which is 12\frac{1}{2}. The equation of the line containing this altitude, with a given yy-intercept of (0,−2)(0, -2), is y=12x−2y = \frac{1}{2}x - 2. Substituting the coordinates of vertex A(k,4)A(k, 4) into the equation gives 4=12k−24 = \frac{1}{2}k - 2. Solving for kk yields k=12k = 12.

Adım Adım Çözüm

1
Calculate the slope of side BCBC using the slope formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}.
mBC=−6−23−(−1)=−84=−2m_{BC} = \frac{-6 - 2}{3 - (-1)} = \frac{-8}{4} = -2
The altitude is perpendicular to the side BCBC, so we first need the slope of BCBC.
2
Find the slope of the altitude by taking the negative reciprocal of the slope of BCBC.
malt=−1mBC=−1−2=12m_{\text{alt}} = -\frac{1}{m_{BC}} = -\frac{1}{-2} = \frac{1}{2}
Perpendicular lines have slopes that are negative reciprocals of each other.
3
Write the equation of the line containing the altitude using the slope-intercept form y=mx+by = mx + b with the given yy-intercept of (0,−2)(0, -2).
y=12x−2y = \frac{1}{2}x - 2
The line has a slope of 12\frac{1}{2} and crosses the yy-axis at −2-2.
4
Substitute the coordinates of vertex A(k,4)A(k, 4) into the equation and solve for kk.
4=12k−2⇒6=12k⇒k=124 = \frac{1}{2}k - 2 \Rightarrow 6 = \frac{1}{2}k \Rightarrow k = 12
Vertex AA lies on the altitude line, so its coordinates must satisfy the line's equation.

Anahtar Kavram

The slope of a line perpendicular to a given line is the negative reciprocal of the given line's slope.

Alternatif Yöntem

Alternatively, we can use the vector dot product. The vector representing side BCBC is BC⃗=(3−(−1),−6−2)=(4,−8)\vec{BC} = (3 - (-1), -6 - 2) = (4, -8). The vector from the yy-intercept to vertex AA is v⃗=(k−0,4−(−2))=(k,6)\vec{v} = (k - 0, 4 - (-2)) = (k, 6). Since the altitude is perpendicular to BCBC, the dot product of these two vectors must equal zero: (4)(k)+(−8)(6)=0⇒4k−48=0⇒4k=48⇒k=12(4)(k) + (-8)(6) = 0 \Rightarrow 4k - 48 = 0 \Rightarrow 4k = 48 \Rightarrow k = 12.
Tahmini Süre:2m 0s
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Konular

Parallel and Perpendicular Lines

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