Write the equation of a line that is perpendicular to the given line and that passes through the given point.

8 y-3= 3(x+2); (-2, 3)
(1 point)
3 Oy+3=-(x-2)
3 Oy-3=-(x+2)
3 Oy-2=-(x+3)
3 Oy+3=(x-2)

1 answer

To find a line that is perpendicular to the given line, we need to determine the slope of the given line. The given line is in slope-intercept form (y = mx + b) where m represents the slope. In this case, the given line's equation is 8y - 3 = 3(x + 2).

Rearranging this equation in slope-intercept form, we get:
8y - 3 = 3x + 6
8y = 3x + 9
y = (3/8)x + 9/8

Therefore, the slope of the given line is 3/8.

A line perpendicular to this line will have a slope that is the negative reciprocal of 3/8, which is -8/3.

Using the slope-intercept form (y = mx + b) and the point (-2, 3) given, we can substitute the values to find the equation of the line.

y = mx + b
3 = (-8/3)(-2) + b
3 = 16/3 + b
b = 3 - 16/3
b = 9/3 - 16/3
b = -7/3

Thus, the equation of the line that is perpendicular to 8y - 3 = 3(x + 2) and passes through the point (-2, 3) is y = (-8/3)x - 7/3.