1. [tex]y+2 = 3(x-2)[/tex]
Point-slope form of the equation of a straight line is:
[tex]y-y_0 = m(x-x_0)[/tex] (1)
The two points in this case are:
[tex](x_0,y_0)=(2,-2)\\(x_1,y_1)=(5,7)[/tex]
Slope of the line is given by:
[tex]m=\frac{y_1 -y_0}{x_1 -x_0}=\frac{7-(-2)}{5-2}=\frac{9}{3}=3[/tex]
Substituting into eq.(1), we find:
[tex]y+2 = 3(x-2)[/tex]
2. [tex]y-4 = \frac{3}{4}(x-6)[/tex]
Point-slope form of the equation of a straight line is:
[tex]y-y_0 = m(x-x_0)[/tex] (1)
The two points in this case are:
[tex](x_0,y_0)=(6,4)\\(x_1,y_1)=(2,1)[/tex]
Slope of the line is given by:
[tex]m=\frac{y_1 -y_0}{x_1 -x_0}=\frac{1-4}{2-6}=\frac{3}{4}[/tex]
Substituting into eq.(1), we find:
[tex]y-4 = \frac{3}{4}(x-6)[/tex]
3. [tex]y+3x=3[/tex]
Standard form of the equation of a straight line is:
[tex]ax+bx=c[/tex]
with a, b, c integer numbers.
Let's start by finding the point slope form first.
Point-slope form of the equation of a straight line is:
[tex]y-y_0 = m(x-x_0)[/tex] (1)
The two points in this case are:
[tex](x_0,y_0)=(0,3)\\(x_1,y_1)=(2,-3)[/tex]
Slope of the line is given by:
[tex]m=\frac{y_1 -y_0}{x_1 -x_0}=\frac{-3-3}{2-0}=-\frac{6}{2}=-3[/tex]
Substituting into eq.(1), we find:
[tex]y-3 = -3(x-0)[/tex]
Now we can re-arrange the equation to re-write it in standard form:
[tex]y-3 = -3(x-0)\\y-3 = -3x\\y-3+3x=0\\y+3x=3[/tex]
4. [tex]y-x=-2[/tex]
Standard form of the equation of a straight line is:
[tex]ax+bx=c[/tex]
with a, b, c integer numbers.
Let's start by finding the point slope form first.
Point-slope form of the equation of a straight line is:
[tex]y-y_0 = m(x-x_0)[/tex] (1)
The two points in this case are:
[tex](x_0,y_0)=(1,-2)\\(x_1,y_1)=(4,2)[/tex]
Slope of the line is given by:
[tex]m=\frac{y_1 -y_0}{x_1 -x_0}=\frac{2-(-1)}{4-1}=\frac{3}{3}=1[/tex]
Substituting into eq.(1), we find:
[tex]y+1 = x-1[/tex]
Now we can re-arrange the equation to re-write it in standard form:
[tex]y+1=x-1\\y+1-x=-1\\y-x=-2[/tex]
