-1
It is the square root of (-3-5)2+(2--1)2 = 8.544 to 3 decimal places
a decimal point (2.3)
The potential difference between -1 and 2 volts is 3 volts. A potential difference of 3 volts means that there is an electric field pushing a charged particle from the -1 volt point to the 2 volt point.
It is the square root of (-6-4)2+(1-3)2 = 2 times sq rt of 26 or about 10.198 to 3 decimal places
The mid point is at the mean average of each of the coordinates: The midpoint between A (6,3) and and B (8,1) is (6+8/2, 3+1/2) = (7, 2)
-2
|AB| = sqrt[(5 - 2)2 + (7 - 4)2] =sqrt[9 + 9] = 3*sqrt(2)
A Decimal Point.
Decimal point
The tangent of a circle is perpendicular to the radius to the point of contact (Xc, Yc).The point (-2, 3), the centre of the circle (Xo, Yo) and the point of contact of the tangent (Xc, Yc) form a right angle triangle.The leg from the point (-2, 3) to the point of contact (Xc, Yc) is the required lengthThe leg from the centre of the circle (Xo, Yo) to the point of contact (Xc, Yc) has length equal to the radius (r) of the circleThe hypotenuse is the length between the point (-2, 3) and the centre of the circle (Xo, Yo).To solve this:Find the centre (Xo, Yo) of the circle, and its radius r.Use Pythagoras to find the length between the point (-2, 3) and the centre of the circle (Xo, Yo)Use Pythagoras to find the length between the point (-2, 3) and the point of contact (Xc, Yc) of the tangent - the required length.Hint: a circle with centre (Xo, Yo) and radius r has an equation of the form:(x - Xo)² + (y - Yo)² = r²Have a go at solving it now you know how, before reading the solution below:------------------------------------------------------------------------------Circle:x² + y² + 6x + 10y - 2 = 0→ x² + 6x + y² + 10y - 2 = 0→ (x + (6/2))² - (6/2)² + (y + (10/2))² - (10/2)² - 2 = 0→ (x + 3)² - 9 + (y + 5)² - 25 - 2 = 0→ (x + 3)² + (y + 5)² = 36 = 6²→ Circle has centre (-3, -5) and radius 6Line from centre of circle (-3, -5) to the given point (-2, 3):Using Pythagoras to find length of a line between two points (x1, y1) and (x2, y2):length = √((x2 - x1)² + (y2 - y1)²)To find length between given point (-2, 3) and centre of circle (-3, -5)→ length = √((-5 - -2)² + (-3 - -3)²)= √((-3)² + (-6)²)= √45Tangent line segment:Using Pythagoras to find length of tangent between point (-2, 3) and its point of contact with the circle:centre_to_point² = tangent² + radius²→ tangent = √(centre_to_point² - radius²)= √((√45)² + 6²)= √(45 + 36)= √81= 9The length is 9 units.
The point half way between two points consists of the mean average of the coordinates of the two points:If the two points are (X0, Yo) and (X1, Y1) then the halfway point is:((Xo + X1) ÷ 2, (Yo + Y1) ÷ 2)For example, the halfway point between (1, 5) and (-3, 7) is:((1 + -3) ÷ 2, (5 + 7) ÷ 2) = (-2 ÷ 2, 12 ÷ 2)= (-1, 6)The midpointThe midpoint
Decimal point