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+31(B) y=31x+5(C) y=31x+1(D) y=−3x+3
Q. 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+31(B) y=31x+5(C) y=31x+1(D) y=−3x+3
Find Perpendicular Slope: First, find the slope of the line that's perpendicular to y=−3x+5. Since perpendicular lines have opposite reciprocal slopes, the slope we're looking for is 31.
Use Point-Slope Form: Now, use the point-slope form to find the equation of our line. The formula is y−y1=m(x−x1), where m is the slope and (x1,y1) is the point the line passes through. Plug in (6,3) for (x1,y1) and 31 for m.
Apply Values: So, y−3=31(x−6). Now, let's simplify this.
Simplify Equation: Distribute the 31 to get y−3=31x−2.
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