Points: (-10, 3) and (-2, -5)
Slope: -1
Equation: y = -x-7
The point-slope form of a line is given by the equation ( y - y_1 = m(x - x_1) ), where ( m ) is the slope and ( (x_1, y_1) ) is a point on the line. Given a slope of -3 and the point (10, -1), we can substitute these values into the formula: ( y - (-1) = -3(x - 10) ). This simplifies to ( y + 1 = -3(x - 10) ), or ( y + 1 = -3x + 30 ). Thus, the point-slope form is ( y + 1 = -3(x - 10) ).
To write the equation of a line in standard form (Ax + By = C) with a slope of 2 that passes through the point (4, -2), we can start with the point-slope form: (y - y_1 = m(x - x_1)). Substituting the point and the slope gives us (y + 2 = 2(x - 4)). Simplifying this, we get (y + 2 = 2x - 8), or rearranging it leads to (2x - y = 10). Thus, the standard form of the equation is (2x - y = 10).
The slope-intercept form of a line is given by the equation ( y = mx + b ), where ( m ) is the slope and ( b ) is the y-intercept. Given a slope ( m = -1 ) and a point (-10, -6), we can substitute these values into the equation to find ( b ): [ -6 = -1(-10) + b \implies -6 = 10 + b \implies b = -16. ] Thus, the slope-intercept form of the line is ( y = -x - 16 ).
If you mean point of (1, 4) and slope of -10 then y = -10x+14
If you mean point of (1, 4) and slope of -10 then y = -10x+14
Point: (2, 4) Slope: -3 Equation: y = -3x+10
To find the point-slope form of a linear equation, you first need the equation in slope-intercept form (y = mx + b) or at least identify a point and the slope. The equation you've provided, "4x y 10," seems incomplete or unclear. If you mean (4x + y = 10), you can rearrange it to (y = -4x + 10), giving a slope (m = -4). The point-slope form is then (y - y_1 = m(x - x_1)) where ((x_1, y_1)) is any point on the line, such as (0, 10).
The point-slope form of a line is given by the equation ( y - y_1 = m(x - x_1) ), where ( m ) is the slope and ( (x_1, y_1) ) is a point on the line. Given a slope of -3 and the point (10, -1), we can substitute these values into the formula: ( y - (-1) = -3(x - 10) ). This simplifies to ( y + 1 = -3(x - 10) ), or ( y + 1 = -3x + 30 ). Thus, the point-slope form is ( y + 1 = -3(x - 10) ).
To write the equation of a line in standard form (Ax + By = C) with a slope of 2 that passes through the point (4, -2), we can start with the point-slope form: (y - y_1 = m(x - x_1)). Substituting the point and the slope gives us (y + 2 = 2(x - 4)). Simplifying this, we get (y + 2 = 2x - 8), or rearranging it leads to (2x - y = 10). Thus, the standard form of the equation is (2x - y = 10).
The slope-intercept form of a line is given by the equation ( y = mx + b ), where ( m ) is the slope and ( b ) is the y-intercept. Given a slope ( m = -1 ) and a point (-10, -6), we can substitute these values into the equation to find ( b ): [ -6 = -1(-10) + b \implies -6 = 10 + b \implies b = -16. ] Thus, the slope-intercept form of the line is ( y = -x - 16 ).
Need point slope form. Y = mX + c 4X + 2Y = 10 2Y = - 4X + 10 Y = - 2X + 5 ========= slope(m) = - 2 -------------------
If you mean point of (1, 4) and slope of -10 then y = -10x+14
Point: (-3, 4) Slope: 2 Equation: y-4 = 2(x--3) => y = 2x+10
If you mean point of (1, 4) and slope of -10 then y = -10x+14
If you mean slope of -10 and point of (1, 4) then the equation is y = -10x+14
If you mean a slope of -10 through the point (1, 4) then the equation is y = -10x+14
Points: (8, 10) and (-4, 2) Slope: 2/3 Equation: 3y = 2x+14