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The equation of the line through (0,2) and (2, 5) is
(y - 2) = (5 - 2)/(2 - 0)*(x - 0)
or 2*(y - 2) = 3*x
or 3x - 2y + 4 = 0
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How do you write an equation of the line that passes through the points. (2 5) (0 5)?
Points: (2, 5) and (0, 5) Slope: 0 Equation: x = 2 meaning that it is a straight horizontal line parallel to the y axis
What are the solutions to the equation 3x2-13x-10 equals 0?
(3x + 2)(x - 5) = 0 x = -2/3 or 5
What is the equation of x equals 2 over 3 and x equals 5 over 7 showing work?
If: x = 2/3 and x = 5/7 Then: (3x-2)(7x-5) = 0 Equation: 21x2-29x+10 = 0
Is 3x4 plus 2x plus 5 equals 0 a quadratic equation?
No. It is a quartic equation. The largest power of x in a quadratic equation must be 2.
What is the equation for The product of 5 and the sum of 8 and 2?
find the sum and product of the roots of 8×2+4×+5=0
3b2 plus b-10 equals 0?
3b2 + b - 10 = 0 This quadratic equation can be factored. (3b -5)(b +2) = 0 The equation can be solved when either (3b - 5) = 0, or (b + 2) = 0. If 3b - 5 = 0 then 3b = 5 : b = 5/3 If b + 2 = 0 then b = -2 The roots or solutions to this equation are b = 5/3 and b = -2
What is the equation of the line that passes through the points 0 5 and 2 3?
Points: (0. 5) and (2, 3) Slope: -1 Equation: y = -x+5
How do you write an equation of the line that passes through the points. (2 5) (0 5)?
Points: (2, 5) and (0, 5) Slope: 0 Equation: x = 2 meaning that it is a straight horizontal line parallel to the y axis
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 are the solutions to the equation 3x2-13x-10 equals 0?
(3x + 2)(x - 5) = 0 x = -2/3 or 5
What is the equation for this line 0-2 20 42?
Points: (0, -2) and (20, 42) Slope: 11/5 Equation: 5y = 11x-10 or as y = 2.2x-2.
What is the equation of the circle that touches the x axis when its center is at 2 5?
Center of circle: (2, 5) Point of contact with the x axis: (2, 0) Distance from (2, 5) to (2, 0) equals 5 which is the radius of the circle Equation of the circle: (x-2)^2 +(y-5)^2 = 25
Y equals -5x plus 5 and 3x plus 2y equals 3 solve this equation using substitution?
y = -5x+5 (Equation 1)3x+2y = 3 (Equation 2)you see the the first equation is equal to y. just put that into the second equation.3x + 2(-5x+5) = 3 (now do algebra and find x)3x - 10x + 10 = 3-7x = -7-x = -1x = 1Now put x into one of the equations to find yy = -5(1) + 5 (Equation 1)y = -5 + 5y = 0Now put both the x and y into both equations to see if it is truee.0 = -5(1) + 5 (Equation 1)0 = -5 + 50 = 0 CORRECT3(1) + 2(0) = 3 (Equation 2)3 + 0 = 33 = 3 CORRECTTherefore your answers arex = 1y = 0
What is the equation of x equals 2 over 3 and x equals 5 over 7 showing work?
If: x = 2/3 and x = 5/7 Then: (3x-2)(7x-5) = 0 Equation: 21x2-29x+10 = 0
What is 2x3 plus 11x equals 6x?
your equation is this... 2x3 + 11x = 6x 2x3 + 5x = 0 x(2x2 + 5) = 0 x = 0 and (5/2)i and -(5/2)i
What is the perpendicular bisector equation of the line whose coordinates are -4 8 and 0 -2 on the Cartesian plane showing work?
Points: (-4, 8) and (0, -2) Slope: (8--2)/((-4-0) = -5/2 Perpendicular slope: 2/5 Midpoint: (-4+0)/2, (8-2)/2 = (-2, 3) Equation: y-3 = 2/5(x--2) Multiply all terms by 5: 5y-15 = 2(x--2) => 5y = 2x+19 Perpendicular bisector equation in its general form: 2x-5y+19 = 0
What is the equation of the line that passes through the points negative 2 2 and 0 5?
Just to be clear, the two points are (-2,2) and (0,5). The slope (rise/run) is (5-2)/(0 - -2) = 3/2. So y = (3/2)*x + b. Put one of the point into this and solve for b: 5 = (3/2)*0 + b; b = 5. The equation is y = (3/2)*x + 5. Do a check that the other point (-2,2) satisfies this equation {and it does}.
