Write an equation in standard form of the line that contains the point (1,-3) and is
a. parallel to the line 8x+3y=5
b. perpendicular to the line 8x+3y=5
a. To find the equation of a line parallel to another line, we need to have the same slope. The given line 8x + 3y = 5 can be rewritten in slope-intercept form as y = -8/3x + 5/3. Therefore, the slope of this line is -8/3. Since a parallel line will have the same slope, the equation we're looking for will have the same slope of -8/3. We can then use the point-slope form of the equation of a line to write the equation. The point-slope form is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.
Plugging in the values for (x1, y1) = (1, -3) and m = -8/3, we get:
y - (-3) = -8/3(x - 1)
y + 3 = -8/3x + 8/3
Multiplying through by 3 to eliminate the fraction, we get:
3y + 9 = -8x + 8
Rearranging the terms to get in standard form, we have:
8x + 3y = -1
b. To find the equation of a line perpendicular to another line, we need to take the negative reciprocal of the slope. The given line 8x + 3y = 5 has a slope of -8/3. Taking the negative reciprocal, we have 3/8. Using the point-slope form of the equation of a line, we can write the equation. Plugging in the values for (x1, y1) = (1, -3) and m = 3/8, we get:
y - (-3) = 3/8(x - 1)
y + 3 = 3/8x - 3/8
Multiplying through by 8 to eliminate the fraction, we get:
8y + 24 = 3x - 3
Rearranging the terms to get in standard form, we have:
-3x + 8y = -27