What is an equation of the line that passes through the point left bracket, minus, 6, comma, 3, right bracket(−6,3) and is perpendicular to the line 6, x, minus, 5, y, equals, 56x−5y=5?

Answer 1

The given line can be rewritten in slope-intercept form as 56x−5y=5. By isolating y, we have -5y = -56x + 5. Dividing by -5, we find y = 11.2x - 1.

Since the line we are looking for is perpendicular to this line, its slope will be the negative reciprocal of 11.2, which is -1/11.2.

Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the point (-6,3) and m is the slope of the line:

y - 3 = -1/11.2(x - (-6))
y - 3 = -1/11.2(x + 6)
y - 3 = -1/11.2x - 6/11.2
y - 3 = -1/11.2x - 15/14
Multiply both sides by 11.2 to eliminate the fraction:
11.2y - 33.6 = -x - 165/14

Finally, rearranging the terms, we obtain the equation of the line perpendicular to the given line and passing through the point (-6,3): x + 11.2y = -165/14 + 33.6.