Question

1. the line represented by the equation 5x - 2y = 10 is transformed by a dilation centered at (2, 0) with a scale factor of 2. the image of the line

a. is the original line
b. passes through the point (4, 0)
c. passes through the point (0, -10)
d. is perpendicular to the original line

1. a line whose equation is y = -2x + 3 is dilated by a scale factor of 4 centered at (0, 3). which equation represents the image of the line after the dilation?

a. y = -2x + 3
b. y = -2x + 12
c. y = -8x + 3
d. y = -8x + 12

Explanation:

Problem 1

Step1: Recall Dilation of Lines

A dilation centered at a point \((h,k)\) with scale factor \(r\) transforms a point \((x,y)\) to \((h + r(x - h), k + r(y - k))\). For a line, if the center of dilation lies on the line, the image of the line under dilation is the same line (since all points on the line, when dilated, still lie on the line as the center is on it). First, check if \((2,0)\) lies on \(5x - 2y = 10\). Substitute \(x = 2\), \(y = 0\): \(5(2)-2(0)=10 - 0 = 10\), which matches the equation. So the center is on the line.

Step2: Analyze Options

• Option A: Since the center of dilation is on the line, dilating the line with any scale factor (here 2) will result in the same line (because all points on the line, when dilated from a point on the line, remain on the line).
• Option B: The point \((4,0)\): Let's see, original line \(5x - 2y = 10\) or \(y=\frac{5}{2}x - 5\). The center is \((2,0)\). A point on the original line: take \(x = 0\), \(y=-5\). Dilating \((0,-5)\) with center \((2,0)\) and scale 2: \(x'=2 + 2(0 - 2)=2-4=-2\), \(y'=0 + 2(-5 - 0)=-10\). So \((-2,-10)\) is on the image. \((4,0)\): Let's check if it's on original line: \(5(4)-2(0)=20

eq10\), so not on original (and image is original, so no).

• Option C: \((0,-10)\): Original line at \(x = 0\) is \(y=-5\), so not on original (image is original, so no).
• Option D: Original line slope is \(\frac{5}{2}\), image is same line, so slopes equal, not perpendicular.

Step1: Recall Dilation of Lines

A dilation centered at \((h,k)\) with scale factor \(r\) transforms a point \((x,y)\) to \((h + r(x - h), k + r(y - k))\). For a line \(y = mx + b\), if the center of dilation \((0,3)\) lies on the line? Check: \(y=-2(0)+3 = 3\), so \((0,3)\) is on the line. So dilating a line with center on the line, the image is the same line (because all points on the line, when dilated from a point on the line, remain on the line).

Step2: Analyze Options

The original line is \(y=-2x + 3\). Since the center \((0,3)\) is on the line, dilation with scale factor 4 will not change the line (same slope, same y-intercept as center is on it).

• Option A: \(y=-2x + 3\) (matches, as image is same line).
• Option B: \(y=-2x + 12\) (different y-intercept, wrong).
• Option C: \(y=-8x + 3\) (different slope, wrong).
• Option D: \(y=-8x + 12\) (different slope and intercept, wrong).

Answer:

A. is the original line

Problem 2