Which equation is the equation of a line that passes through (–10, 3) and is perpendicular to y = 5x – 7?
(1 point)
Responses
y = 5x + 53
y = –one fifthx – 7
y = –one fifthx + 1
y = one fifthx + 5
To find the equation of a line that is perpendicular to y = 5x - 7, we need to find its slope and then take the negative reciprocal of that slope.
The slope of y = 5x - 7 is 5.
The negative reciprocal of 5 is -1/5.
Now, we can use the point-slope form of the equation of a line to find the equation that passes through (-10, 3) with a slope of -1/5.
Using the point-slope form:
y - y1 = m(x - x1)
where (x1, y1) is (-10, 3) and m is -1/5, the equation becomes:
y - 3 = -1/5(x - (-10))
y - 3 = -1/5(x + 10)
y - 3 = -1/5x - 2
y = -1/5x - 2 + 3
y = -1/5x + 1
Therefore, the equation of the line that passes through (-10, 3) and is perpendicular to y = 5x - 7 is:
y = -1/5x + 1