It is a straight line equation that can be expressed as y = 1/3x+1
Perpendiculat straight lines.
3x - 2y = -2 and 6x - 4y = 0Solving both equations for y, we havey = (3/2)x + 1 and y = (3/2)xSince both lines have the same slope, they are parallel lines.
The locus of points equidistant from lines y = 0 and x = 3 is the line y = -x + 3.
They are perpendicular lines because the slopes are 3/4 and -4/3 respectively.
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parallel
Perpendiculat straight lines.
perpendicular
3x - 2y = -2 and 6x - 4y = 0Solving both equations for y, we havey = (3/2)x + 1 and y = (3/2)xSince both lines have the same slope, they are parallel lines.
The locus of points equidistant from lines y = 0 and x = 3 is the line y = -x + 3.
They are perpendicular lines because the slopes are 3/4 and -4/3 respectively.
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If the second equation is: y minus 2x equals 3, then:y - 2x = 3 ⇒ y = 2x + 3 and it is parallel to y = 2x.Otherwise (with with missing operator as "plus", "multiply" or "divide"), the lines are neither parallel nor perpendicular.
They are intersecting lines.
x^3y = 2, y = 4 Substitute 4 for y: (x^3)(4) = 2 x^3 = 8 x = 2
The lines intersect at (3, 5)
That depends on the other equation which has not been given but if 2x-3y = 2 then y = 2/3x-2/3