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What is the equation of the line graphed in the 
xy-plane that passes through the point 
(6,3) and is perpendicular to the line whose equation is 
y=-3x+5 ?
Choose 1 answer:
(A) 
y=3x+(1)/(3)
(B) 
y=(1)/(3)x+5
(C) 
y=(1)/(3)x+1
(D) 
y=-3x+3

What is the equation of the line graphed in the xy x y -plane that passes through the point (6,3) (6,3) and is perpendicular to the line whose equation is y=3x+5 y=-3 x+5 ?\newlineChoose 11 answer:\newline(A) y=3x+13 y=3 x+\frac{1}{3} \newline(B) y=13x+5 y=\frac{1}{3} x+5 \newline(C) y=13x+1 y=\frac{1}{3} x+1 \newline(D) y=3x+3 y=-3 x+3

Full solution

Q. What is the equation of the line graphed in the xy x y -plane that passes through the point (6,3) (6,3) and is perpendicular to the line whose equation is y=3x+5 y=-3 x+5 ?\newlineChoose 11 answer:\newline(A) y=3x+13 y=3 x+\frac{1}{3} \newline(B) y=13x+5 y=\frac{1}{3} x+5 \newline(C) y=13x+1 y=\frac{1}{3} x+1 \newline(D) y=3x+3 y=-3 x+3
  1. Find Perpendicular Slope: First, find the slope of the line that's perpendicular to y=3x+5y=-3x+5. Since perpendicular lines have opposite reciprocal slopes, the slope we're looking for is 13\frac{1}{3}.
  2. Use Point-Slope Form: Now, use the point-slope form to find the equation of our line. The formula is yy1=m(xx1)y - y_1 = m(x - x_1), where mm is the slope and (x1,y1)(x_1, y_1) is the point the line passes through. Plug in (6,3)(6,3) for (x1,y1)(x_1, y_1) and 13\frac{1}{3} for mm.
  3. Apply Values: So, y3=13(x6)y - 3 = \frac{1}{3}(x - 6). Now, let's simplify this.
  4. Simplify Equation: Distribute the 13\frac{1}{3} to get y3=13x2y - 3 = \frac{1}{3}x - 2.

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