Write an equation of the line that passes through the point P
and is perpendicular to the line with the given equation:
P(5,3); y = 5x + 2
I think its Y=1/5x+4
nope. The slope must be -1/5
So the equation would be Y=-1/5 + 4?
Not quite...
The slope is -1/5
So the new equations is y= -1/5 (x) + b
now sub in the point (5,3) and solve for b
and then you will have the new equation.
So it would be Y= -1/5x + 28?
To find the equation of a line that is perpendicular to another line, we need to remember that the slopes of perpendicular lines are negative reciprocals of each other.
The given equation of the line is y = 5x + 2. We can see that the slope of this line is 5 (the coefficient of x).
To find the slope of the line perpendicular to this line, we'll take the negative reciprocal of 5. The negative reciprocal of 5 is -1/5.
Now we have the slope (-1/5) and the point (5,3) that lies on the line we want to find the equation for.
We'll use the point-slope form of a linear equation, where the equation is y - y1 = m(x - x1). Here, (x1, y1) represents the coordinates of the point and m represents the slope.
Plugging in the values, we get:
y - 3 = (-1/5)(x - 5)
Now we simpllify and convert the equation to slope-intercept form (y = mx + b):
y - 3 = (-1/5)x + 1
y = (-1/5)x + 1 + 3
y = (-1/5)x + 4
Therefore, the equation of the line that passes through point P(5,3) and is perpendicular to y = 5x + 2 is y = (-1/5)x + 4.