Find an equation of the line containing the given pair of points (1,5)and(3,6)

The "slope intercept form of the equation of a (the) line" is y=mx+b, where m is the slope of the line and b is the y-intercept.

We are given the slope of 1/2, so m= 1/2.

We can now write y=1/2 x + b.

Since the point (-2,-3) is on the line, we can substitute it in and solve for b. We put the -2 in for x and -3 in for y.

-3 = 1/2(-2) +b
-3 = -1 + b
-3 + 1 = -1 + b +1
-2 =b

Thus, the equation of the line is y= 1/2 x -2

To check if we did this correctly, plug in the point (-2, -3) to see if it works.
Left Side Right Side
-3 = 1/2 (-2) -2
= -1-2
= -3

9x-5y=121

Being parallel to the given line, the equation of the line you are seeking has the same slope, which in this case is a=1/4.
So the equation sought is as follows
y=(1/4)x +b, where b is to be found.
To find b, use the stated fact that the line passes through the point (x=8, y=-1). All that means is that the point (8,-1) is on the line whose equation you are looking for. If it is on the line with equation y=(1/4)x+b
then its coordinates x=8, and y=-1 satisfy the relation y=(1/4)x+b. In other words, if you substitute 8 for x, and -1 for y, the equality holds true -1=(1/4)*8 +b
This gives you a way to find the initial value of the function (the y-intercept b ). Just solve -1=(1/4)*8 +b to find b.
I leave this pleasure to you.

1. No solutions: The system is incoherent, incompatible Example: 2x+3y=8 and 2x+3y= 15. The two lines are parallel and distinct.
2. One solution: There exits a pair of values (x,y) that satisfy both linear equations. The two lines on a Cartesian graph have one intersection point.
3. Infinite number of solutions: The two equations are one and the same (one is just multiplied by some constant). The graph of the two lines yields the same line. One is superposed on the other. Any ordered pair (x,y) that satify one equation (there is an infinity of such pairs) satisfies the other.
4. Two solutions: cannot happen because the two lines can either intersect once, be parallel, or superposed one on the other.
   

1. Transform this equation to its functional form:
2. 9y=-3x+17 or y=(-1/3)x+17/9
3. In the last equation, the slope is the coefficient of x, namely -1/3.
   
4. A line parallel to this one must have the same slope (-1/3).
5. So the equation of your line starts this way: y=(-1/3)x+b.
6. To identify (calculate) b, you must make use of the fact that the parallel line passes through the point (1,5).
   
7. That means that the coordinates of the point (1,5) satisfy the equation of the parallel line y=(-1/3)x+b
8. Substitute 5 for y, and 1 for x and solve for b.

And that is as far I will go. I leave it to you finish up the work : find b and write the equation in the functional form, then convert it to the general form (Ax+By+C=0) if that is what you are asked to produce.

You are looking for a line (y=m*x+b) and have two points. From this information you can generate two equations with two unknowns (m and b are unknown).
First plug in the first point (1,5) to the general form:
5(the y value of the point) = m*1(the x-value at this point)+b
Do the same for the second point you're given.
From here solve the first equation for m in terms of b.
Plug this value of 'm' into the second equation so you will end up with something like:
3=(something in terms of b)*(-2)+b
This final equation can be solved for b (try factoring)
You now have a value for the y-intercept. Plug that into y=m*x+b
Choose either of the two points, plug into the equation on the last line with the value of b known
You then know y, x, and b and have m as the remaining 1 unknown. Solve for that and put it all together for your final answer.

The site seems to eat the plus signs I enter, so I will use PLUS to symbolize addition.

To find the equation of the straight line (y = a*x PLUS b) that passes through two points P1(x1,y1) and P(x2,y2) , you need to use
1. the coordinates of the points to calculate the slope a (gradient) as a=(y2-y1)/(x2-x1)
2. Replace the calculated value of a in the equation and write that one of the points ( P1(x1,y1) for example) satisfies the equation. In other words y1=a*x1 PLUS b.
Here y1 and x1 are known values, a has been calculated, and only b is still unknown. You can now use the equation y1=a*x1 PLUS b to calculate b as
b=(y1-a*x1)

Example: Equation of the line through (1,5) and (3,6)
Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5, x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2
The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=(1/2)*3 PLUS b.
Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5
Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

I trust you can substitute you own values for (x1,y1, x2,y2) to duplicate the calculations above.

Example: Equation of the line through (1,5) and (3,6)
Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5, x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2
The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=3/2 PLUS b.
Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5
Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=1, y1=0, x2=0, and x1=-6. You should get a=(1-0)/(0-(-6))=1/6
The y-intercept is the y-cordinate for x=0. Its value is 1.
The equation is then y=(x/6) 1.

assuming the question is what is the circle equation?
and if (-2,2) is the center of the circle
the equation should look like this: (x+2)^2+(Y-2)^2=R^2

And now only R is needed.

given 2x-5y+4=0 equation of line perpendicular

we can rearange the equation to be y=(2x+4)/5
from that we can see that the slope of the line is 2/5
And from the fact of perpendicular line we can say that the slope
of the radius line is -2/5.

The motivation now is to calculate the distance between the center of the circle to the cross point of the radius with the line perpendicular

For that we would calculate the radius line equation and compare it to the equation of line perpendicular

As mentioned earlier the slope of the radious line is -2/5.

So the equation is y=-2/5x+b and b can be calculated by using the center of the circle coordinates

2= - (2/5)*(-2)+b ------> b=2-4/5=1.2
radius equation is y=-(2/5)x+1.2

Now the cross point is calculated by comparing the equations:
-(2/5)x+1.2=(2x+4)/5 --> -2x+6=2x+4 --> 4x=2 --> x=1/2 --> y=1

So the cross point is (1/2,1).

The distance between the points is calculated by the following
Formula:

R=SQR(((1/2)-(-2))^2+(2-1)^2)=SQR(2.5^2+1^2)=SQR(6.25+1)=
SQR(7.25)

Therefore the circle eq is (x+2)^2+(Y-2)^2=7.25