Question

friday

1. find the slope between the given points: (-2, 5) and (6, -10)
2. simplify: $\frac{2x^{2}y^{3}}{3xy^{2}}$
3. find the coordinates of the image given the dilation around the origin by a factor of 3.
4. given the following transformation, write the rule:

Explanation:

Step1: Recall slope formula

Slope $m = \frac{y_2 - y_1}{x_2 - x_1}$

Step2: Substitute points $(-2,5),(6,-10)$

$m = \frac{-10 - 5}{6 - (-2)} = \frac{-15}{8}$

Step1: Separate coefficients/variables

$\frac{2}{3} \cdot \frac{x^2}{x} \cdot \frac{y^3}{y^2}$

Step2: Simplify using exponent rules

$\frac{2}{3} \cdot x^{2-1} \cdot y^{3-2} = \frac{2xy}{3}$

Step1: Identify pre-image coordinates

Pre-image points: $N(2,1), J(3,-3), T(4,-2), K(5,2)$

Step2: Dilate each coordinate by 3

$N'(2 \cdot 3, 1 \cdot 3)=(6,3)$; $J'(3 \cdot 3, -3 \cdot 3)=(9,-9)$; $T'(4 \cdot 3, -2 \cdot 3)=(12,-6)$; $K'(5 \cdot 3, 2 \cdot 3)=(15,6)$

Step1: Compare pre-image/image points

Pre-image: $A(-4,3), B(-3,-3), C(1,3), D(0,1)$; Image: $A'(4,3), B'(3,-3), C'(-1,3), D'(0,1)$

Step2: Determine coordinate transformation

$x$-coordinate flips sign, $y$-coordinate stays: $(x,y) \to (-x,y)$

Answer:

1. $\boldsymbol{-\frac{15}{8}}$
2. $\boldsymbol{\frac{2xy}{3}}$
3. $N'(6, 3)$, $J'(9, -9)$, $T'(12, -6)$, $K'(15, 6)$
4. $\boldsymbol{(x, y) \to (-x, y)}$