note, x-axis at a, therefore coordinates is (a, 0).
y-axis at b, therefore (0, b)
a) gradient of I can be found by converting 3x-4y=24 into eqn of y=mx+c (just some simple manipulation.)
3x-4y=24
-4y=-3x+24
4y=3x-24
( ÷ by 4)
y=(3/4)x-6
gradient is hence, 3/4.
b) length of AB,
substitute y=0 into equation of line I to find point a, and separately x=0 to find point b.
once you've found the full coordinates, you may insert the values into this formula to find length of line segment.
[tex] \sqrt{(y1 - y2) + (x1 - x2)} [/tex]
if my workings are not wrong, you should find that a=8 and b=-6 and the length is 10 units.
c) equation of line passing through b, grad = OM
find gradient OM, which is
[tex] \frac{ - 3 + 0}{ 4 - 0} = \frac{ - 3}{4} [/tex]
from this, substitute points b (0,-6) into
y=mx + c in order to find the 'c' constant. once, you've found the c, simply insert that c value along with the gradient of OM into y=mx+c
therefore,
[tex]y = - \frac{3}{4} x - 6[/tex]
