Question

A. The point-slope form of the equation of a line is y ? y1 = m(x ? x1), where m is the slope and (x1, y1) is a point on the line. Write the equation of the line in point-slope form perpendicular to the graph of y = 1/2x -7 passing through the point (6, 5).

B. Write an equation of the perpendicular bisector of JK, where J = (?8, 4) and K = (4, 4).

Can you show work please.

Answer 1

A.

`y - 5 = -2(x-6)`

Negative reciprocal gives you the perpendicular slope so negative reciprocal of 1/2 is -2.

Then apply point-slope form.

B. The answer is x = 6.

The midpoint of JK is

`\left( (x_1 + x_2)/(2), (y_1 + y_2)/(2) \right) = \left( (8+ 4)/(2), (4 + 4)/(2) \right) = \left(6,4\right)`

The line that goes through JK is just a horizontal line
`y = 4` because the y-coordinate does not change. So its perpendicular bisector is the vertical line that goes through the x-coordinate of the midpoint, that is,
`x = 6`.