Determine the equations of a straight line passing through (-6;-2) and pependicular to 3x+2y=6
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1. First, you need to find the slope of the given line (3x+2y=6).
2. Then, once you find the slope, determine the perpendicular slope.
3. Once you have the perpendicular slope, determine the equation for a line which goes through (-6, -2).
If you have any questions about any of these steps, please feel free to reply to this thread.
Thanks!
Leo
P(-6,-2).
3x + 2y = 6
m1 = -A/B = -3/2
m2 = -(-2/3) = 2/3 = Negative reciprocal
Y = mx + b = -2
2x/3 + b = -2
(2*-6)/3 + b = -2
-4 + b = -2
b = 2.
Eq: Y = 2x/3 + 2
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To find the equation of a line that is perpendicular to another line, we need to make use of the fact that the slopes of perpendicular lines are negative reciprocals of each other.
First, let's determine the slope of the given line: 3x + 2y = 6. We need to rewrite this equation in slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept.
Rearranging the equation:
2y = -3x + 6
y = (-3/2)x + 3
From this equation, we can conclude that the slope of the given line is -3/2.
To find the slope of the line perpendicular to this, we take the negative reciprocal of -3/2. The negative reciprocal of a number is obtained by flipping the fraction and changing the sign. So, the slope of the perpendicular line is 2/3.
Now, let's find the equation of the line passing through the point (-6, -2) with a slope of 2/3.
Using the point-slope form of a linear equation, which is y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is a point on the line, we substitute the given values into the equation:
y - (-2) = 2/3(x - (-6))
y + 2 = 2/3(x + 6)
y + 2 = 2/3x + 4
y = 2/3x + 2
Therefore, the equation of the straight line passing through (-6, -2) and perpendicular to 3x + 2y = 6 is y = (2/3)x + 2.