equation for linear functions

Question

line passing through (1,2) and perpendicular to the graph g(x)=-1/6x+1

Answer 1

To find this line, let's consider the second part of it being perpendicular to g(x) = (-1/6)x + 1

For this line to be perpendicular to g(x) we must identify the slope of g(x) and then take the negative reciprocal of it. That will be the slope of the line.

For g(x) = (-1/6)x + 1, it's in the y = mx + b format, where m is the slope and b is the y-intercept. From comparison, m = -1/6

The slope of our line will be the negative reciprocal of this, so we'll have m = 6 after exchanging places with the numerator and denominator and multiplying it by -1

With m = 6 as the slope for the line, we can use the point (1,2) which it passes through to determine the y-intercept for the line. Since y = mx + b implies y = 6x + b for our line, we can plug the x and y values of the point (1,6) into this last equation and solve for b

y = 6x + b

2 = 6(1) + b

2 = 6 + b

b = -4

Since we have b = -4 and m = 6, the line which is perpendicular to g(x) = (-1/6)x + b and that passes through (1,2) is

y = 6x - 4

 

Hope this helps!