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The image shows line 
p. For each equation below, decide whether the line it represents is parallel to line 
p, perpendicular to line 
p, or neither of these.

y=-3x+10
neither

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

y=(1)/(3)x
[Select]

-3x+y=1
[Select]

The image shows line \newlinepp. For each equation below, decide whether the line it represents is parallel to line \newlinepp, perpendicular to line \newlinepp, or neither of these.\newliney=3x+10y=-3x+10\newlineneither\newliney5=(13)(x+2)y-5=-(\frac{1}{3})(x+2)\newline[Select]\newliney=13xy=\frac{1}{3}x\newline[Select]\newline3x+y=1-3x+y=1\newline[Select]

Full solution

Q. The image shows line \newlinepp. For each equation below, decide whether the line it represents is parallel to line \newlinepp, perpendicular to line \newlinepp, or neither of these.\newliney=3x+10y=-3x+10\newlineneither\newliney5=(13)(x+2)y-5=-(\frac{1}{3})(x+2)\newline[Select]\newliney=13xy=\frac{1}{3}x\newline[Select]\newline3x+y=1-3x+y=1\newline[Select]
  1. Determine Slope of Line: Determine the slope of line pp from the given equation y=3x+10y = -3x + 10. Comparing this with the standard form y=mx+by = mx + b, we identify the slope (mm) of line pp as 3-3.
  2. Analyzing Equation y5y - 5: Analyze the equation y5=13(x+2)y - 5 = -\frac{1}{3}(x + 2). First, simplify it to slope-intercept form: y=13x132+5y = -\frac{1}{3}x - \frac{1}{3}\cdot2 + 5, which simplifies to y=13x+413y = -\frac{1}{3}x + 4 \frac{1}{3}. The slope here is 13-\frac{1}{3}. Since the slopes 3-3 and 13-\frac{1}{3} are not equal nor opposite reciprocals, this line is neither parallel nor perpendicular to line pp.
  3. Consider Equation y=(13)xy = (\frac{1}{3})x: Consider the equation y=(13)xy = (\frac{1}{3})x. The slope here is 13\frac{1}{3}. Since 13\frac{1}{3} is not equal to 3-3 and also not the negative reciprocal of 3-3, this line is neither parallel nor perpendicular to line pp.
  4. Examine Equation 3x+y-3x + y: Examine the equation 3x+y=1-3x + y = 1. Rearrange it to slope-intercept form: y=3x+1y = 3x + 1. The slope here is 33. Since 33 is not equal to 3-3 and is not the negative reciprocal of 3-3, this line is neither parallel nor perpendicular to line pp.

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