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max2010maxim [7]

2 years ago

11

Quadrilateral abcd is transformed to create A’B’c’d’ match the coordinates of A’ with the transformations that create it

1 answer:

notka56 [123]2 years ago

6 0

Answer:

its easy!

Step-by-step explanation:

Answer:Quadrilateral ABCD is transformed to create A′B′C′D′. Match the coordinates of A′ with the transformations that create it. (2, 4) (2, 1) (6, 12) (4, -5) (-2, 4) (2, -4) (6, -12) (-4, 5) Quadrilateral ABCD is reflected over the x-axis. arrowRight Quadrilateral ABCD is translated 2 units right and 1 unit down. arrowRight Quadrilateral ABCD is dilated by a scale factor of 3. arrowRight Quadrilateral ABCD is rotated 180° clockwise about the origin.

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Answer:

Parallel line:

$y=-\frac{4}{5}x+\frac{9}{5}$

Perpendicular line:

$y=\frac{5}{4}x-\frac{1}{2}$

Step-by-step explanation:

we are given equation 4x+5y=19

Firstly, we will solve for y

$4x+5y=19$

we can change it into y=mx+b form

$5y=-4x+19$

$y=-\frac{4}{5}x+\frac{19}{5}$

so,

$m=-\frac{4}{5}$

Parallel line:

we know that slope of two parallel lines are always same

so,

$m'=-\frac{4}{5}$

Let's assume parallel line passes through (1,1)

now, we can find equation of line

$y-y_1=m'(x-x_1)$

we can plug values

$y-1=-\frac{4}{5}(x-1)$

now, we can solve for y

$y=-\frac{4}{5}x+\frac{9}{5}$

Perpendicular line:

we know that slope of perpendicular line is -1/m

so, we get slope as

$m'=\frac{5}{4}$

Let's assume perpendicular line passes through (2,2)

now, we can find equation of line

$y-y_1=m'(x-x_1)$

we can plug values

$y-2=\frac{5}{4}(x-2)$

now, we can solve for y

$y=\frac{5}{4}x-\frac{1}{2}$

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3 years ago