write an equation passing through the point and perpendicular to the given line (6,1); y = -3/5x+7

Question

write an equation passing through the point and perpendicular to the given line (6,1); y = -3/5x+7

Answer 1

Any line perpendicular to the given line will have a slope of 5/3. This is because perpendicular lines have slopes that are negative reciprocals, i.e. the product of their respective slopes equals -1.

Now you want to write the equation of a line passing through (6, 1) having a slope of 5/3. There are a few ways to determine that equation.

One way is to use the point-slope formula: y - y₁=m(x - x₁), where m is the slope and (x₁,y₁) is the ordered pair for the point the line passes through.

y - 1 = (5/3)(x - 6).

To convert to slope-intercept form:

y - 1 = (5/3)x - 10

y = (5/3)x - 10 + 1

y = (5/3)x - 9

Or you can start with slope-intercept: y = mx + b where m is the slope and (x, y) = (6,1).

1 = (5/3)(6) + b

1 = 10 +b

b = 1 - 10

b = -9

Once you have determined the value for b: y = (5/3)x - 9

Visit this graph to see that the green line is always perpendicular to the red line because the slopes are negative reciprocals.

desmos.com/calculator/lmdc9o1u4x