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What is the equation of the line that passes through the point (4,11) and is perpendicular to the line with the equation [tex]\( y = \frac{4}{3}x + 7 \)[/tex]?

A. [tex]\( y = -\frac{3}{4}x + 8 \)[/tex]
B. [tex]\( y = \frac{4}{3}x + 7 \)[/tex]
C. [tex]\( y = \frac{4}{3}x - 15 \)[/tex]
D. [tex]\( y = -\frac{3}{4}x + 14 \)[/tex]


Sagot :

To determine the equation of the line that passes through the point [tex]\((4,11)\)[/tex] and is perpendicular to the line given by the equation [tex]\(y = \frac{4}{3}x + 7\)[/tex], we can follow these steps:

1. Identify the slope of the given line:
The equation [tex]\(y = \frac{4}{3} x + 7\)[/tex] is in slope-intercept form [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope. Therefore, the slope of the given line is [tex]\(\frac{4}{3}\)[/tex].

2. Determine the slope of the perpendicular line:
The slope of a line perpendicular to another line is the negative reciprocal of the slope of the original line. Hence, we need to find the negative reciprocal of [tex]\(\frac{4}{3}\)[/tex]:
[tex]\[ m_{\perp} = -\frac{1}{m} = -\frac{1}{\frac{4}{3}} = -\frac{3}{4} \][/tex]
Therefore, the slope of the perpendicular line is [tex]\(-\frac{3}{4}\)[/tex].

3. Use the point-slope form to find the equation of the perpendicular line:
The point-slope form of the equation of a line is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\(m\)[/tex] is the slope of the line and [tex]\((x_1, y_1)\)[/tex] is a point on the line. For our perpendicular line, [tex]\(x_1 = 4\)[/tex], [tex]\(y_1 = 11\)[/tex], and [tex]\(m_{\perp} = -\frac{3}{4}\)[/tex]. Substituting these values, we get:
[tex]\[ y - 11 = -\frac{3}{4}(x - 4) \][/tex]

4. Simplify the equation:
To put it in slope-intercept form ([tex]\(y = mx + b\)[/tex]), we distribute and simplify:
[tex]\[ y - 11 = -\frac{3}{4}x + \frac{3}{4}(4) \][/tex]
[tex]\[ y - 11 = -\frac{3}{4}x + 3 \][/tex]
Adding 11 to both sides, we get:
[tex]\[ y = -\frac{3}{4}x + 14 \][/tex]

Therefore, the equation of the line that passes through the point [tex]\((4, 11)\)[/tex] and is perpendicular to the line [tex]\(y = \frac{4}{3}x + 7\)[/tex] is [tex]\(y = -\frac{3}{4}x + 14\)[/tex].

The correct answer is:
[tex]\[ \boxed{D. \, y = -\frac{3}{4} x + 14} \][/tex]
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