Answer:
1) [tex]\mathsf{ y+3x-9=0}}[/tex]
2) [tex]\mathsf{ y+x=0}}[/tex]
Step-by-step explanation:
The eqn. of a line passing thru two given points, (x₁, y₁) and (x₂, y₂):
[tex]\boxed{\mathsf{(y-y_1)=\frac{y_2-y_1}{x_2-x_1} (x-x_1)}}[/tex]
The given points are (1, 6) and (2, 3), take any one of them as (x₁, y₁), but remember your choice for the eqn. changes if you make a mistake.
I'll be substituting (1, 6) in place of (x₁, y₁)
[tex]\mathsf{\implies (y-6)=\frac{3-6}{2-1} (x-1)}[/tex]
[tex]\mathsf{\implies (y-6)=\frac{-3}{1} (x-1)}[/tex]
[tex]\mathsf{\implies (y-6)=-3(x-1)}[/tex]
[tex]\mathsf{\implies y-6=-3x+3}[/tex]
Taking all of them to the left side (signs get reversed in doing so):
[tex]\mathsf{\implies y-6+3x-3=0}[/tex]
Operating on like terms:
[tex]\mathsf{\implies y+3x-(6+3)=0}[/tex]
[tex]\mathsf{\underline{\implies y+3x-9=0}}[/tex]
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JUSTIFICATION:
(1, 6) lies on the line y + 3x - 9 =0, so it must satisfy it's equation:
Substituting x and y with 1 and 6 respectively:
=> 6 + 3(1) - 9 = 0
=> 6 + 3 - 9 = 0
=> 9 - 9 = 0
=> 0 = 0
True! Hence, the point (1, 6) lies on the line y + 3x - 9 = 0
The given points are:
Substituting (-1, 1) in place of (x₁, y₁):
[tex]\mathsf{\implies (y-1)=\frac{[-1]-1}{1-[-1]} (x-[-1])}[/tex]
two negative signs when placed together make a positive:
[tex]\mathsf{\implies (y-1)=\frac{-1-1}{1+1} (x+1)}[/tex]
[tex]\mathsf{\implies (y-1)=\frac{-2}{2} (x+1)}[/tex]
[tex]\mathsf{\implies (y-1)=-1 (x+1)}[/tex]
[tex]\mathsf{\implies y-1=-1 x-1}[/tex]
Taking all the terms to the left (signs get reversed):
[tex]\mathsf{\implies y \:\overline{-1+1}+x=0}[/tex]
canceling (-1) and (+1):
[tex]\mathsf{\underline{\implies y+x=0}}[/tex]
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JUSTIFICATION:
(-1, 1) lies on the line y + x =0, so it must satisfy it's equation:
Substituting x and y with -1 and 1 respectively:
=> -1 + 1 = 0
=> 0 = 0
True! Hence, the point (-1, 1) lies on the line y + x = 0
