gradient and distance

[vague veldt]

i really need help completing the question, i dont know what numbers i need to square root

[unborn python]

can you find the coordinates of A and C?

[vague veldt]

sorry i sent the wrong one

a (-2, 11)
c (2, 4)

[unborn python]

right, so since the the y coordinate of A is 11, the distance between the x axis and point A is 11,
similarly the distance between the x axis at point C is 4. are you comfortable with that?

[vague veldt]

yep

do i square root and square those

[unborn python]

well yeah, if you just want the formula u just do sqrt((y coordinate of A - y coordinate of C)^2 + (x coordinate of A - x coordinate of C)^2)

[fringe pewter]

,, d=\sqrt{(y_A-y_C)^2+(x_A-x_C)^2}

[lavish sand]

רי-ית

[vague veldt]

so
square root (11-4)^2 + (-2-2)^2

[unborn python]

yep

[vague veldt]

alright let me try it

8.06

thank youu i got it right

[unborn python]

nw :D

[quartz lake]

Notice that you are making use of pythagoras'

[silent violet]

If you are done with this channel, please mark your problem as solved by typing .close

[fair cobalt]

!done

[silent violet]

If you are done with this channel, please mark your problem as solved by typing .close

[vague veldt]

.close