Can someone help answer this?

in a right angled triangle, (height)^2 + (base)^2 = (hypotenuse)^2

here we have , height as X and base as X too , so; x^2 + x^2 = (9√2)^2

2(X)^2 = 162

x^2 = 162/2

X^2 = 81

X = √81 = 9

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like,   however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

so the negative reciprocal of the slope of ef is 1, however the slope of df is not 1, is 1/2, so nope, they're not perpendicular lines.

the answer to the question is a

step-by-step explanation:

it has the same slope and the same y-intercept.

y + 11 = -7/9(x - 18)

y + 11 =-7/9x + 14

b

y = -7/9x + 3

y-intercept (0, 3)