Question

The solution of two linear equations is (-2,2). One equations has a slope of 3. The slope of the other equation is the negative reciprocal of the slope of the first.
The system described above is represented by the following equations.

Answer 1

The system of equations is:

$3x-y=-8$

$x+3y=4$

Step-by-step explanation:

Let the line 1 be: $l_1$

Let line 2 be: $l_2$

Both lines will pass through the same point (-2,2)

And

Given

$m_1=3\\m_2=-\frac{1}{3}$

The slope intercept form for first line will be:

$y=m_1x+b$

Putting values

$y=3x+b$

Putting the point in the equation

$2=(-2)(3)+b\\2=-6+b\\b=2+6\\b=8$

So, the equation is: y=3x+8

Converting the equation in standard form

$y=3x+8$

Subtracting y from both sides

$0=3x-y+8$

Subtracting 8 from both sides

$3x-y=-8$

The second line's equation will be:

$y=m_2x+b$

Putting slope

$y=-\frac{1}{3}x+b$

Putting the point

$2=-\frac{1}{3}(-2)+b\\2=\frac{2}{3}+b\\b=2-\frac{2}{3}\\b=\frac{6-2}{3}\\b=\frac{4}{3}$

The equation is:

$y=-\frac{1}{3}x+\frac{4}{3}$

Converting the equation in standard form

Multiplying whole equation by 3

$3y=-x+4$

Adding x on both sides

$x+3y=4$

Hence

The system of equations is:

$3x-y=-8$

$x+3y=4$

Keywords: Equation of line, Linear equation

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