Find an equation for a line that contains (4, 9) and is perpendicular to the graph of y = 7x + 3/8
Answer choices
1. y = -1/7 x + 9 4/7
2. y = 7x - 15
3. y = 1/7x + 8 5/7
No work needed.
Since you know that the slopes of perpendicular lines must be negative reciprocals of each other, all you have to look at is their slope.
the slope of the given line is 7, so the slope of the new line must be -1/7
Only 1. fits that.
( I would still check if the given point satisfies y = (-1/7)x + 9 4/7 , it does )
thanks
To find the equation of a line that is perpendicular to the given line, we need to determine the slope of the given line, and then find the negative reciprocal of that slope.
The given equation is y = 7x + 3/8. We can see that the slope of this line is 7, as it is the coefficient of x.
To find the negative reciprocal of the slope, we take the reciprocal of 7 (1/7) and then multiply it by -1. So, the slope of the line perpendicular to the given line is -1/7.
Now that we have the slope (-1/7), we can use the point-slope form of a linear equation to determine the equation of the line. The point-slope form is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line, and m is the slope.
We are given the point (4, 9) that the line must pass through. So, substituting the values into the point-slope form, we have:
y - 9 = (-1/7)(x - 4)
To simplify this equation, we distribute -1/7 to the terms inside the parentheses:
y - 9 = (-1/7)x + 4/7
Next, we isolate y by adding 9 to both sides of the equation:
y = (-1/7)x + 4/7 + 63/7
Combining the fractions on the right side:
y = (-1/7)x + 67/7
Thus, the equation of the line that is perpendicular to the graph of y = 7x + 3/8 and passes through the point (4, 9) is y = (-1/7)x + 67/7.
None of the answer choices provided match this equation, so the correct answer is not among the given options.