Question

Write an equation for the line perpendicular to y = 2x – 5 that contains (–9, 6)

Answer 1

The line perpendicular to y = 2x – 5 passing through (�9, 6) has a slope of -1/2. Using the point-slope form, the equation is y - 6 = -1/2(x + 9), which simplifies to y = -1/2x + 1.5 in slope-intercept form.

Step-by-step explanation:

To find the equation of the line that is perpendicular to y = 2x – 5 and passes through the point (–9, 6), first, we need to understand the relationship between the slopes of perpendicular lines. If a line has a slope m, then any line perpendicular to it will have a slope that is the negative reciprocal, –(1/m). The given line y = 2x – 5 has a slope of 2, so our perpendicular line must have a slope of –(1/2).

We can use the point-slope form equation of line, which is y – y₁ = m(x – x₁), where (x₁, y₁) is a point on the line (in this case, –9, 6) and m is the slope.

Plugging in our values, we get:

y – 6 = –½(x + 9)

This is the perpendicular line's equation in point-slope form. To convert it to slope-intercept form y = mx + b, we distribute the –½ and solve for y:

y – 6 = –½x – –½(9)

y = –½x + –4.5 + 6

y = –½x + 1.5

Therefore, the equation of our perpendicular line in slope-intercept form is y = –½x + 1.5.

Answer 2

Write an equation for the line
perpendicular to y = 2x - 5 that contains (-9,6).
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The given line has slope = 2
Any line perpendicular to it must have slope = (-1/2)
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Form:
y = mx + b
Solve for "b":
6 = (-1/2)(-9) + b
6 = (9/2) + b

b = 3/2
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Equation:
y = (-1/2)x + (3/2)