The system of equations is:
[tex]3x-y=-8[/tex]
[tex]x+3y=4[/tex]
Step-by-step explanation:
Let the line 1 be: [tex]l_1[/tex]
Let line 2 be: [tex]l_2[/tex]
Both lines will pass through the same point (-2,2)
And
Given
[tex]m_1=3\\m_2=-\frac{1}{3}[/tex]
The slope intercept form for first line will be:
[tex]y=m_1x+b[/tex]
Putting values
[tex]y=3x+b[/tex]
Putting the point in the equation
[tex]2=(-2)(3)+b\\2=-6+b\\b=2+6\\b=8[/tex]
So, the equation is: y=3x+8
Converting the equation in standard form
[tex]y=3x+8[/tex]
Subtracting y from both sides
[tex]0=3x-y+8[/tex]
Subtracting 8 from both sides
[tex]3x-y=-8[/tex]
The second line's equation will be:
[tex]y=m_2x+b[/tex]
Putting slope
[tex]y=-\frac{1}{3}x+b[/tex]
Putting the point
[tex]2=-\frac{1}{3}(-2)+b\\2=\frac{2}{3}+b\\b=2-\frac{2}{3}\\b=\frac{6-2}{3}\\b=\frac{4}{3}[/tex]
The equation is:
[tex]y=-\frac{1}{3}x+\frac{4}{3}[/tex]
Converting the equation in standard form
Multiplying whole equation by 3
[tex]3y=-x+4[/tex]
Adding x on both sides
[tex]x+3y=4[/tex]
Hence
The system of equations is:
[tex]3x-y=-8[/tex]
[tex]x+3y=4[/tex]
Keywords: Equation of line, Linear equation
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