How do I write an equation of a line perpendicular to the graph of
-2y + x =-5 that passes through the point at (1,2)
Given the line -2y + x =-5, solve for y so the equation is in this form:
y = mx + b
Once you know the slope, m, you also know the slope of the perpendicular line.
As a reminder, the slope of the perpendicular line will be the negative reciprocal, or -1/m, m being the slope of the line above.
XY 3x+2y=-6,z(3,2)
To start, let's rearrange the given equation -2y + x = -5 to the form y = mx + b, where m represents the slope.
To do this, we need to isolate the term with y on one side of the equation.
-2y + x = -5
We can begin by moving the x term to the right side by adding 2y to both sides of the equation:
x = 2y - 5
Next, we need to isolate the term with y. We can do this by subtracting x from both sides:
2y = x - 5
Finally, we can isolate y by dividing both sides by 2:
y = (1/2)x - 5/2
Now that we have the equation in the form y = mx + b, we can see that the slope of the given line is 1/2.
To find the slope of a line perpendicular to this line, we take its negative reciprocal. The negative reciprocal of 1/2 is -2/1 or simply -2.
Now that we have the slope of the perpendicular line, we can use the given point (1,2) to find the equation of the line.
We can use the slope-intercept form (y = mx + b) and substitute the values of the slope (-2), the x-coordinate (1), and the y-coordinate (2) into the equation:
2 = -2(1) + b
Simplifying, we have:
2 = -2 + b
Now we can solve for b by adding 2 to both sides:
b = 4
Therefore, the equation of the line perpendicular to the graph of -2y + x = -5 that passes through the point (1,2) is:
y = -2x + 4.