Write the standard form of the equation of the line through the given point with the given slope.
through: (3, 5), slope = 5/3
I'm using the point-slope formula of y-y1=m(x-x1)
y-y1=m(x-x1)
y-5=5/3(x-3)
y-5=5/3x-15/3
y-5=5/3x - 5
y-5+5=5/3x -5+5
y=5/3x
This is where I am stuck. Not sure what to do from here. Did I plug in the xs and ys incorrectly?
I figured it out. Nevermind. Sorry to bother y'all.
Had to do y=5/3x + 0
then do 3*y=3*5/3x + 0
to get,3y=15/3x + 0
3y=5x + 0
3y-5x=5x-5x + 0
Answer: 5x-3y=0
(3, 5), m = 5/3.
Y = mx + b = 5.
(5/3)3 + b = 5,
b = 0.
Y = (5/3)x + 0.
Standard form: (5/3)x - Y = 0.
You have correctly applied the point-slope formula to the given information. However, there seems to be a small mistake in one step. Let's go through the process together.
Starting with the point-slope form: y - y1 = m(x - x1)
Given point: (3, 5)
Slope: 5/3
Plug in the values:
y - 5 = (5/3)(x - 3)
Next, we can simplify the right side:
y - 5 = (5/3)x - 5
Now, add 5 to both sides of the equation to isolate y:
y - 5 + 5 = (5/3)x - 5 + 5
This simplifies to:
y = (5/3)x
So the final equation in standard form is:
3y = 5x
You have correctly used the point-slope formula to find the equation of the line. However, there seems to be a mistake in your calculation while simplifying the equation.
Starting from the equation: y - 5 = (5/3)(x - 3)
To simplify, distribute (5/3) to (x - 3):
y - 5 = (5/3)x - (5/3)3
y - 5 = (5/3)x - 15/3
y - 5 = (5/3)x - 5
Next, add 5 to both sides to isolate y:
y - 5 + 5 = (5/3)x - 5 + 5
y = (5/3)x
So, the standard form of the equation of the line is y = (5/3)x. This form represents a line with slope 5/3 passing through the point (3, 5).