I need help with the bottom question

Step-by-step explanation:

they want to know  a point on the line... that has a gradient ( think slope )  of -1/3  and has the point of (-2,5)  on it. So find the formula for that line

with the point slope formula  y-y1 = m(x - x1)

plug in what we know

y - 5 = - 1/3 (x -(-2))

y - 5 = -1/3(x+2)

y - 5 = -1/3x - 2/3

y = - 1/3x - 2/3 + 5

y = - 1/3x + 4 1/3

y = -1/3x + 13 / 3

Now use your formula to find any new point plug in 1 for x ( b/c it's easy)

y = 12/3   when x= 1.  so ( 1, 4 )  is another point on the line  :)

don't forget to show your work  :P

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option

don't give

take it once

try and try and you will not

happy to

pls mark as brainliest