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Gradient of line = (4 - 1)/(2 + 3) = 3/5
So general form of equation: y = 3x/5 + c where c is a constant
or 5y = 3x + c' where c' = 5c is a constant
The point (2,4) is on this line
so 20 = 6 + c' so tha c' = 14
Therefore, the equation of the line is: 5y = 3x + 14
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What is the equation and its perpendicular bisector equation of the line joining the points of 1 2 and 3 4 showing work?
1 Points: (1, 2) and (3, 4) 2 Slope: (2-4)/(1-3) = 1 3 Perpendicular slope: -1 4 Midpoint: (1+3)/2 and (2+4)/2 = (2, 3) 5 Equation: y-2 = 1(x-1) => y = x+1 6 Bisector equation: y-3 = -1(x-2) => y = -x+5
What is the equation of a line in general form that passes through points (-1 2) and (5 2)?
Points: (-1, 2) and (5, 2) Slope: 0 Equation: y = 2
What is the equation of the straight line through the points -1 6 and 2 -6?
Points: (-1, 6) and (2, -6) Slope: -4 Equation: y = -4x+2
What is the straight line equation joining the points of 7 0 and 0 2?
It is in its general form: 2x+7y-14 = 0
The equation of the line joining through -1 3 amd parallel to the line joining 6 3 2 -3?
To find the equation of a line parallel to another line, we need the same direction vector. The direction vector of the given line is (2, -3). Therefore, the equation of the line parallel to it passing through (-1, 3) is x = -1 + 2t and y = 3 - 3t, where t is a parameter.
What is the equation and its perpendicular bisector equation of the line joining the points of 1 2 and 3 4 showing work?
1 Points: (1, 2) and (3, 4) 2 Slope: (2-4)/(1-3) = 1 3 Perpendicular slope: -1 4 Midpoint: (1+3)/2 and (2+4)/2 = (2, 3) 5 Equation: y-2 = 1(x-1) => y = x+1 6 Bisector equation: y-3 = -1(x-2) => y = -x+5
What is the equation of the line joining the points 1 1 and 7 2?
Points: (1, 1) and (7, 2) Slope: (2-1)/(7-1) = 1/6 Equation: y -1 = 1/6(x -1) y = 1/6x -1/6+1 => y = 1/6x +5/6 Which can be expressed in the form of: x -6y +5 = 0
What is the equation of the line passing through the points 2 2 and 3 1?
Points: (2, 2) and (3, 1) Slope: -1 Equation: y = -x+4
What is the slope-intercept equation of the line that contains the points (1 2) and (0 -2)?
Points: (1, 2) and (0, -2) Slope: 4 Equation: y = 4x-2
What is the equation of a line in general form that passes through points (-1 2) and (5 2)?
Points: (-1, 2) and (5, 2) Slope: 0 Equation: y = 2
What equation of the line that contains the point 1 -1 and 2 -2?
Choose the equation of the line that contains the points (1, -1) and (2, -2).
What is the equation of the straight line through the points -1 6 and 2 -6?
Points: (-1, 6) and (2, -6) Slope: -4 Equation: y = -4x+2
What is the straight line equation joining the points of 7 0 and 0 2?
It is in its general form: 2x+7y-14 = 0
What is the equation of the line passing through the given points 1 and 5 and 2 and 7?
Points: (1, 5) and (2, 7) Slope: 2 Equation: y = 2x+3
The equation of the line joining through -1 3 amd parallel to the line joining 6 3 2 -3?
To find the equation of a line parallel to another line, we need the same direction vector. The direction vector of the given line is (2, -3). Therefore, the equation of the line parallel to it passing through (-1, 3) is x = -1 + 2t and y = 3 - 3t, where t is a parameter.
What is the equation of the line through points -15 and 2-4?
If you mean points of: (-1, 5) and (2, -4) Then the equation works out as: y = -3x+2
What is the equation of the line passing through the points 0 2 and 60?
Points: 0 2 and 6 0 Equation: y = -1/3x+2
