y=1/3 x -6, y equals one third x minus 6
The equation will look like -5/3 - -8/-6 since we're finding the slope. That equals 3/9 or 1/3, the slope is 1/3. But since we also need to find the y-intercept, we need to solve another equation, y-y1=m(x-x1). Now we just need to put in the values. y-(-5)=1/3(x-3) this becomes y+5=1/3 x - 1, subtract 5 to isolate y, y=1/3 x -6.
Any coordinate with a y value of 12 lies on the line y = 12.
Any coordinate with an x value of -5 lies on the line x = -5.
Since the equation must be perpendicular to the line x = 4, which is a vertical line, it must also be written in the form x = a.
Since the x coordinate is 3, the equation of the line is simply x = 3.
slope of line perpendicular to x=4 is 0
reqd. eq. of line is y=-5
The answer is
To find an equation of a line given the slope and a point we use the formula
m is the slope
( x1 , y1) is the point
From the question the point is (3,-5) and slope 0
We have the final answer as
Hope this helps you
y = -⁵/₄x -⁵/₄
The point-slope formula for a straight line is
y₂ - y₁ = m(x₂ - x₁)
The line goes through the points (3, -5) and (-5,5).
5 – (-5) = m(-5 – 3)
10 = -8m
m = -⁵/₄
y = mx + b
Use the point (5, -5)
-5 = -⁵/₄×5 + b
-5 = -²⁵/₄ + b
b = ⁵/₄
∴ y = -⁵/₄x - ⁵/₄
y=-5/4 + -1.25
y1 - y2 over x1 -x2
5- -5 =10
simplify: -10/8 which is -5/4 *THAT IS YOUR SLOPE*
Now to find the y intercept pick a point. I chose (-5,5) and plug in for y=mx+b
y will be 5 ; m is your slope so -5/4 ; x is -5 and we don't have out b yet.
5=-5/4(-5) + b
multiply -5/4 and -5 = 6.25
then just subtract -5 and -6.25 = -1.25 which is your y intercept
y= -5/4 + -1.25
hope this helps :3
if it did pls mark brainliest
Put it in the form y=mx+b=>y=2x-1/2. Instead of -1/2, put it as b, since you want it to go through 3,-5. y=2m+b when it goes through (3,-5) happens when -5=3*2-b. Do algebra, and get b as 11. That means the equation is y=2x-11.
You didn't provide the slope, but the generic equation is
In your case, , so we have
Substitute with the slope you want and you have the equation of the line