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The x-intercept of the line that passes through the intersection of the lines x + 2y = 4 and 2x + 3y = 6, and is perpendicular to the line 3x — y = 2 is
The x-intercept of the line that passes through the intersection of the
lines x + 2y = 4 and 2x + 3y = 6, and is perpendicular to the line 3x — y = 2 is
Let us find the intersection point of the lines
x + 2y = 4 and 2x + 3y = 6
Multiplying (x + 2y = 4) by 2 and subtracting with 2x + 3y = 6
2x+4y = 8
2x+3y = 6
y = 2
So, x = 0
Therefore, the point of intersection is (0,2).
We know that a line that passes through (0,2) is perpendicular to the line 3x-y = 2
The slope of the line 3x-y = 2
y = 3x-2
Slope is 3
We know the product of slopes of perpendicular lines is $$m_1m_2=-1$$
Therefore, the line perpendicular to 3x - y = 2 will have a slope of -1/3
Thus, the line passing through (0,2) with a slope of -1/3 will be
$$y-2\ =-\dfrac{1\left(x-0\right)}{3}$$
3y-6 = -x
3y+x = 6
Thus, the x-intercept of this line will be when y = 0
so, x = 6
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