Using Pythagoras:
distance = √(change_in_x2 + change_in_y2)
= √((5 - -8)2 + (4 - 4)2)
= √(132 + 02)
= √(132)
= 13 units.
When y=4, x=8/5
4/5 - 1/2 = 8/10 - 5/10 =(8-5)/10 =3/10 =0.3 (point three)
Using the distance formula the length of ab is 5 units
8 times 4 is 32, but more to the point here, it is 4 eights. 8 times 5 is 40, but more to the point here, it is 5 eights. Obviously then, 5 eights is 1 eight more than 4 eights. Or 8. (and if you don't believe me, subtract 32 from 40 and see if I was right!)
5/8 / 4/7 = 5/8 * 7/4 = 5*7 / 8*4 = 35 / 32
If you mean points of (5, 4) and (5, -4) then the distance works out as 8
If you mean: (-8, 4) and (5, 4) Then the distance between the points works out as 13
7.62
To find the perpendicular distance from the point ( (4, 8) ) to the line ( y = 2x + 10 ), we first rewrite the line in standard form: ( 2x - y + 10 = 0 ). The formula for the distance ( d ) from a point ( (x_0, y_0) ) to a line ( Ax + By + C = 0 ) is given by: [ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} ] Substituting ( A = 2 ), ( B = -1 ), ( C = 10 ), ( x_0 = 4 ), and ( y_0 = 8 ): [ d = \frac{|2(4) - 1(8) + 10|}{\sqrt{2^2 + (-1)^2}} = \frac{|8 - 8 + 10|}{\sqrt{4 + 1}} = \frac{10}{\sqrt{5}} = 2\sqrt{5} \approx 4.47 ] Thus, the perpendicular distance from the point ( (4, 8) ) to the line is approximately ( 4.47 ).
If you mean points of: (2, 5) and (-4, 8) Distance is the square root of (2--4)^2+(5-8)^2 = 6.708 rounded
3
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 There is a distance of 12.
5x/8+10y/13=-3/4
When y=4, x=8/5
Fractions that are equal to .5 are 5/10, 1/2, 2/4, 4/8, etc.
4/5 - 1/2 = 8/10 - 5/10 =(8-5)/10 =3/10 =0.3 (point three)
If you mean points of (2, 4) and (-1, 8) then the distance works as 5