The point of intersection is when a set value of x and y works in both equations - this is a set of simultaneous equations to solve:
solving (3) gives x = 5.5
Substituting for y in (2) gives 5y - 3 × 5.5 = 6 → 5y = 22.5 → y = 4.5
Checking by substituting for y and x in (1) gives 4×4.5 - 2×5.5 = 18 - 11 = 7 as required
→ the point of intersection of 4y - 2x = 7 and 5y - 3x = 6 is (5.5, 4.5)
It works out that the tangent line of y -3x -5 = 0 makes contact with the circle of x^2 + y^2 -2x +4y -5 = 0 at (-2, -1)
Lines: 4y-2x = 7 and 5y-3x = 6 Multiply the 1st line by 5 and the 2nd line by 4 So: 20y-10x = 35 and 20y-12x = 24 Subtract the 1st line from the 2nd line: 2x = 11 => x = 11/2 Through substitution both lines intersect at: (11/2, 9/2)
It is 2√5 ≈ 4.47 units.To solve this:Find the equation of the line perpendicular to y = 2x + 10 that passes through the point (2, 4);Find the point where this line meets y = 2x + 10Find the distance from this point to (2, 4) using PythagorasThe slope of the perpendicular line (m') to a line with slope m is such that mm' = -1, ie m' = -1/mFor y = 2x + 10, the perpendicular line has slope -1/2, and so the line that passes through (2, 4) with this slope is given by:y - 4 = -½(x - 2)→ 2y - 8 = -x + 2→ 2y + x = 10To find where this meets the line y = 2x + 10, substitute for y in the equation of the perpendicular line and solve for x:y = 2x + 102y + x = 10→ 2(2x + 10) + x = 10→ 4x + 20 + x = 10→ 5x = -10→ x = -2Now use one of the equations to solve for y:y = 2x + 10→ y = 2(-2) + 10→ y = -4 + 10 = 6This the perpendicular line from (2, 4) meets the line y = 2x + 10 at the point (-2, 6)The distance between these two points is given by:distance = √((-2 - 2)² + (6 - 4)²) = √(16 + 4) = √20 = 2√5 ≈ 4.47 units
8
The minimum value of the parabola is at the point (-1/3, -4/3)
The intersection is (-2, 6)
3
They intersect at the point of: (-3/2, 11/4)
y = 2x - x2y = 0Since both quantities on the right side are equal to 'y', they're equal to each other.2x - x2 = 0x (2 - x) = 0x = 0and2 - x = 0x = 2The two points of intersection are (0, 0) and (2, 0) .
2X-Y=3 X+Y=3 ---------- 3X = 6 X=2 2(2)-Y=3 4-Y=3 Y=1 Point of interesection: (2,1).
There are two equations in the question, not one. They are the equations of intersected lines, and their point of intersection is their common solution.
Solving the simultaneous equations works out as x = -2 and y = -2 So the lines intersect at: (-2, -2)
They intersect when 2x + 4 = -x + 1 or 3x = -3 => x = -1 and when x = -1, y = -x + 1 = -(-1) + 1 = 2 So the point of intersection is (-1, 2)
y = 5y = -2x + 3Since both of those things are equal to 'y', they're equal to each other.5 = -2x + 3Add 2x to each side:5 + 2x = 3Subtract 5 from each side:2x = -2Divide each side by 2:x = -1The point of intersection is (-1, 5) .
Perpendicular equation: x+2y = 0 Point of intersection: (2, -1) Perpendicular distance: square root of 5
If: y = 3x^2 +10x +11 and y = 2 -2x -x^2 Then: 3x^2 +10x +11 = 2 -2x -x^2 Transposing terms: 4x^2 +12X +9 = 0 Factorizing the above: (2x+3)(2x+3) = 0 meaning x = -3/2 By substitution into original equations intersection is at (-3/2, 11/4)
If you mean a line of y = 2x+5/4 and a curve of y^2 = 10x Then it works out that the line touches the curve at: (5/8, 5/2)