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

First, we need to find the slope of the given line. The equation of the line is in the form Ax + By = C. We can rewrite this in the slope-intercept form (y = mx + b) to find the slope.

The given line is 6x - 5y = 5. Let's rewrite this:

-5y = -6x + 5 y = (6/5)x - 1

So, the slope (m1) of the given line is 6/5.

The slope of the line perpendicular to this line is the negative reciprocal of m1. So, the slope (m2) of the required line is -5/6.

Now, we know that the equation of a line is y = mx + b. We know m2 and we know that the line passes through the point (-6,3). We can substitute these values into the equation to find b:

3 = (-5/6)(-6) + b 3 = 5 + b b = 3 - 5 b = -2

So, the equation of the line that passes through the point (-6,3) and is perpendicular to the line 6x - 5y = 5 is y = -5/6x - 2.