PLZ HELP

On a coordinate plane, triangle J K L has points (0, negative 1), (negative 2, 1), and (3, 3).

Reflect the triangle over the y-axis and find the coordinate of the image of point K.

K’(2, 1)

K’(–2, 1)

K’(–2, –1)

K’(2, –1)

K'(2, 1 )

Step-by-step explanation:

Under a reflection in the y- axis

a point (x, y ) → (- x, y ), thus

K(- 2, 1 ) → K'(2, 1 )

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?