y = 2x - x2
y = 0
Since both quantities on the right side are equal to 'y', they're equal to each other.
2x - x2 = 0
x (2 - x) = 0
x = 0
and
2 - x = 0
x = 2
The two points of intersection are (0, 0) and (2, 0) .
x-y = 1 => x = y+1 x2+y2 = 5 => (y+1)(x+1)+y2 = 5 2y2+2y-4 = 0 y = -2 or y = 1 So the points of intersection are: (-1, -2) and (2, 1)
The points of intersection of the equations 4y^2 -3x^2 = 1 and x -2 = 1 are at (0, -1/2) and (-1, -1)
If: y = x^2 +3x -10 and y = -x^2 -8x -15 Then: x^2 +3x -10 = -x^2 -8x -15 Transposing terms: 2x^2 +11x +5 = 0 Factorizing the above: (2x +1)(x +5) = 0 meaning x = -1/2 or -5 Therefore by substitution points of intersection are at: (-1/2, -45/4) and (-5, 0)
point of crossing y axis is where x = 0 y = 5x -2 let x = 0 y = -2, point of crossing y axis
it equals 0. 0 x 0 will always be 0
It works out that the point of intersection is at (-4, -3.5) on the Cartesian plane.
2
The point of intersection of the given simultaneous equations of y = 4x-1 and 3y-8x+2 = 0 is at (0.25, 0) solved by means of elimination and substitution.
(4, -7)
It works out that they intersect at: (4, -7)
By a process of elimination and substitution the lines intersect at: (1/4, 0)
If: y = x2+20x+100 and x2-20x+100 Then: x2+20x+100 = x2-20x+100 So: 40x = 0 => x = 0 When x = 0 then y = 100 Therefore point of intersection: (0, 100)
By a process of elimination and substitution the lines intersect at: (4, -7)
Perpendicular equation: x+2y = 0 Point of intersection: (2, -1) Perpendicular distance: square root of 5
Equation: y = 8x^2 -26x+15 Equation when factorized: y = (4x-3)(2x-5) When x = 0 then y = (0, 15) which is the point of intersection on the y axis When y = 0 then x = (3/4, 0) and (5/2, 0) which are the points of intersection on the x axis
They are (0, 0)They are (0, 0)They are (0, 0)They are (0, 0)
It is the origin, with coordinates (0, 0). It is normally denoted by O.