Question

An equation for this question? Perpendicular p(2,8) y=2x+1

Answer 1

The equation y = 2x + 1 represents a line with a slope of 2 and a y-intercept of 1. The point (2,8) satisfies this equation.

Step-by-step explanation:

The equation for a straight line in the form y = mx + b can be used to represent a line on a graph. In this equation, m represents the slope of the line, and b represents the y-intercept, which is the point where the line crosses the vertical y-axis.

For the given equation y = 2x + 1, the slope m is 2 and the y-intercept b is 1. This means that for every increase of 1 in the horizontal x-axis, the line will rise by 2 units on the vertical y-axis.

Using the given point p(2,8), we can substitute the x and y values into the equation to check if they satisfy the equation. Plugging in x = 2 and y = 8 into y = 2x + 1, we get 8 = 2(2) + 1. This equation is true, so the point lies on the line.

Answer 2

y = 2x + 1....the slope here is 2. A perpendicular line will have a negative reciprocal slope. To find the negative reciprocal, u flip the slope and change the sign. So the slope is 2 or 2/1......flip it and it becomes 1/2....change the sign and it becomes -1/2. So ur perpendicular line will have a slope of -1/2.

y = mx + b
slope(m) = -1/2
(2,8)....x = 2 and y = 8
now sub into the formula and find b, the y int
8 = -1/2(2) + b
8 = -1 + b
8 + 1 = b
9 = b
so ur perpendicular equation is : y = -1/2x + 9 <====

ur y int is (0,9)....to find the x int, sub in 0 for y and solve for x
0 = -1/2x + 9
1/2x = 9
x = 9 * 2
x = 18...so ur x int is (18,0)...but ur graph doesn't go this far

to graph....ur y int (where the line crosses the y axis) is at (0,9)..and since u have a slope of -1/2, u come down 1 space, and to the right 2 spaces...plot that....then down 1, and to the right 2...plot that....keep doing this and u would cross the x axis at (18,0) if it went that far.....then just connect ur plotted points and u have ur line