write the equation of each line with the given properties, in slope intercept form?

Question

1) through (6,3) and (5,2)
2)through (-6,-2) and (5,-3)

show how you did it too, i am trying to figure this stuff out.
thanks!

Answer 1

2-3 / 5-6
m= -1/-1=1
y=mx+b
3=1(6)+b
3=6+b
b= -3
y=x-3

-3-(-2) / 5-(-6)
m= -1/11
y=mx+b
-2= -1/11(-6)+b
-2=6/11+b
b= -28/11
y= -1/11x-28/11

Answer 2

Slope-intercept style is y = mx+b, the place m is the slope and b is the intercept. The "intercept" is the cost of y while x = 0. So for the 1st question, you have given m=-3 and a value for y (0) and x(4). So replace: 0 = -3*4+b Simplify and you recognize b. For the 2d equation, you're able to desire to push the equation around till you get it to appear like y=mx+b. subsequently, that's exceptionally elementary: subtract 3x from the two area and you get: y = 4 - 3x in case you rearrange the words somewhat greater you would be wanting the slope and y intercept. i will pass away the final one to you, because of the fact i think you typed it in incorrect; there is not any y time era in it. that's no longer certainly needed to have one, however. do you recognize what meaning?

Answer 3

The Equation is like: y=mx+b and:

m = (3-2) / (6-5) = 1/1 = 1

and:
y=mx+b
y=1x+b
=>
3=1(6)+b
b = -3

Equation: y = x-3
_____________________________________________
2)
m = (-2-(-3)) / (-6-5) = -1/11

and:
y=mx+b
y=(-1/11)x+b
=>
-3=(-1/11)(5)+b
b = 5/11-3 = 5/11 - 33/11 = -28/11

Equation: y = -(x+28)/11

₢

Answer 4

The canonic equation of any line connecting 2 points A=[x1,y1] and B=[x2,y2] is:
(y-y1) / (y2-y1) = (x-x1) / (x2-x1) or
(x2-x1) * (y-y1) = (y2-y1) * (x-x1);
thus (1): (5-6)*(y-3)=(2-3)*(x-6) or y-3=x-6 or y-x+3=0;
thus (2): (5+6)*(y+2)=(-3+2)*(x+6) or 11y+22=-x-6 or 11y+x+28=0;