What is the equation of the pair of points (-5,-8) and (-3, -1)

Assuming the 'standard form' is "slope-intercept", calculate the slope from the equation m = y2-y1 = 5 - 1 = 4 = -2
x2-x1 4 - 6 -2
The intercept can be found by substituting either of the two points into the equation y = mx + b
5 = (-2)4 + b
5 = (-8) + b
13 = b
(OR, using the other point, y = mx + b
1 = (-2)6 + b
1 = (-12) + b
13 = b )
Then expressing in general:
y = (-2) x + 13

Both equations are straight lines with different slopes and different intercepts. Where these two lines cross is the ordered pair (x,y) required. This point is (0.15, 0.115)

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.
   

Calcualte the slope of the line as
a=(7-(-11))/(10-(-5))=18/15=6/5
Use the fact that the line passes through one of the two points, for example (10,7)
7=(6/5)*10+b=12+b
Obtain b as b=7-12=-5
The equation of the line in functional form is y=(6/5)x-5
Multiply everything by 5 to clear the fraction
5y=6x-25 or 0=6x-5y-25
Finally, the equation in general form (standard?) is 6x-5y-25=0.

Check the calculation by verifying that the point (10,7) lies on the line.
6(10)-5(7)-25=60-35-25=60-60=0 CHECKed!
Check that the second point (-5,-11) lies on the line also (if you want to)
6*(-5)-5*(-11)-25=-30+55-25=0
That checks OK.

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=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.

6x- 3 = y; (0, -3)

Here 6x - 3 = y is the equation
The ordered pair is (0, -3 )
It means if you substitutes 0 for x and -3 for y the equation will be true
Let us substitute and see:-
'6x' (means 6 times x) means 6 times 0 which is equal to 0
So the equation 6x - 3 = y becomes
0- 3 = -3 is correct
The ordered pair means the values of x and y which make the equation correct
Hope this is clear enough
Good luck
luciana44

2y - x = 3
x = 3y - 5

Add the two equations side by side,

2y - x + x = 3 + 3y - 5

2y = 3y - 2

y = 2

Plug this in the second equation to get x,

x = 3(2) - 5

x = 1

So the solution is x = 1, y = 2

or in ordered pair notation (1, 2)

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