perpendicular JEW
5
2x + 5y = 16 + (-5x) - 2y = 2 That gives 2x + 5y = 2 . . . . . . . . . . (I) and -5x - 2y = - 14 or 5x + 2y = 14 . . . . (II) (I)*5: 10x + 25y = 10 (II)*2: 10x + 4y = 28 Subtracting the second from the first, 21y = -18 so y = -18/21 = -6/7 and then by (I) x = 1 - 5y/2 = 1 - 5/2*(-6/7) = 1 + 15/7 = 22/7 So the ordered pair is (22/7, -6/7)
If: 3x+2y = 5x+2y = 14 Then: 3x+2y = 14 and 5x+2y =14 Subtract the 1st equation from the 2nd equation: 2x = 0 Therefore by substitution the solutions are: x = 0 and y = 7
7+4xz+52xz+2y=7+56xz+2y
With a line in the form y = mx + c, it has gradient m and the line perpendicular to it has gradient m' such that mm' = -1, ie m' = -1/m. A line through a point (x0, y0) with gradient m' has an equation of: y - y0 = m' (x - x0) which can be rearranged to a form for y = mx + c. Thus for the line 2x - 3y = 7: 2x - 3y = 7 → 3y = 2x + 7 → y = 2/3 x + 7/3 → it has gradient 2/3 → perpendicular line has gradient -3/2 → perpendicular line through (4, 9) perpendicular to 2x - 3y = 7 has equation: y - 9 = (-3/2)(x - 4) → 2y - 18 = -3(x - 4) → 2y - 18 = -3x + 12 → 2y = 30 - 3x → 3x + 2y = 30
x + 2y = 7 2y = -x + 7 y = -(x-7)/2 => -x/2 + 7/2 2x - y = 7 -y = -2x + 7 y = 2x + 7 Since -1/2 is the negative reciporical of 2, the slopes of these equations are perpendicular. Therefore, these two lines are perpendicular.
They are parallel lines.
The slope of the line of 2x plus 2y equals 7 is (7/2x - 1).
y
Multiply First equation by 2: 6y - 2x = 0Add second equation to this: 6y - 2x + 2x + 2y = 0 + 7Simplify: 8y = 7 so y = 0.875 and x = 2.625
If: 3x+2y = 5x+2y = 7 Then: 3x+2y = 7 and 5x+2y = 7 Subtract the 1st equation from the 2nd equation: 2x = 0 or x = o By substitution: x = 0 and y = 3.5
5
This form. Y = mX + c- 2X - 7Y = - 2- 7Y = 2X - 2Y = (- 2/7)X + 2/7==============Y intercept = 2/7----------------------
Sum Numb!
No. They are parallel.
(-1, 5)
(3, 2)