Question

i don’t understand how to do question c for the first problem, and for the second question what is a gradient and how do i find it? I need to show work, can someone please help me

Answer 1

1a. Use the Pythagoren theorem:

AC^2 = 7.2^2 + 9.6^2 ==> AC = 12

1b. Same as in (1a), but now use AC as the length of another leg, and AG would be the hypotenuse in triangle ACG:

AG^2 = AC^2 + 3.5^2 ==> AG = 12.5

1c. This requires some trigonometry. In the triangle ACG, we have two legs of length AC = 12 and CG = AE = 3.5. Then the angle $\theta$ made by AG with the floor, which is the same as the angle made by AG and AC, is such that

$\tan\theta=\dfrac{CG}{AC}=\dfrac{3.5}{12}\implies\theta=\tan^{-1}\dfrac{3.5}{12}\approx16.26^\circ$

2a. (6, -5) is not on the line because

$3(-5)+2(6)=-15+12=-3\neq9$

2b. Gradient is another word for slope. Here we can rewrite the line as

$3y+2x=9\implies y=-\dfrac23x+3$

which tells us the slope is -2/3.

2c(i). Any line perpendicular to L has a slope equal to the negative reciprocal of the slope of L. Here, that would be 3/2.

2c(ii). Using the point-slope formula, this perpendicular line has equation

$y+5=\dfrac32(x-6)\implies y=\dfrac32x-14\implies3x-2y=28$