For example, say your given equation is 5y−x=10{\displaystyle 5y-x=10}. To isolate y{\displaystyle y}, first move −x{\displaystyle {-x}} to the opposite side of the equation by adding it to both sides to get 5y=x+10{\displaystyle 5y=x+10} Get rid of the 5{\displaystyle 5} on the 5y{\displaystyle 5y} by dividing both sides of the equation by 5{\displaystyle 5}. The new equation will be y=15x+2{\displaystyle y={\frac {1}{5}}x+2}.
Remember that m{\displaystyle m} represents the slope of the line. The opposite reciprocal of the equation y=15x+2{\displaystyle y={\frac {1}{5}}x+2} would be −51x{\displaystyle {-}{\frac {5}{1}}x} or −5{\displaystyle -5}.
Remember that b{\displaystyle b} represents the y-intercept of the line. For example, say your given point is (8,2){\displaystyle (8,2)} where 8{\displaystyle 8} represents the x{\displaystyle x} coordinate and 2{\displaystyle 2} is the y{\displaystyle y} coordinate. Replace the letters in the y=mx+b{\displaystyle y=mx+b} equation with your known values of slope and xy coordinates: 2=−5(8)+b{\displaystyle 2={-5}(8)+b}
To isolate b{\displaystyle b} in the equation 2=−40+b{\displaystyle 2={-40}+b}, add 40{\displaystyle {40}} to both sides. The equation for the y-intercept of the perpendicular line will be 42=b{\displaystyle 42=b}
The formula for the perpendicular line would come out to be y=−5x+42{\displaystyle y={-5}x+42}
For example, you may be asked to find the coordinates of a line that passes through (6,1){\displaystyle (6,1)} based on a line that passes through (4,6){\displaystyle (4,6)} and (2,3){\displaystyle (2,3)}. Focus on (4,6){\displaystyle (4,6)} and (2,3){\displaystyle (2,3)} for now.
If your points are (4,6){\displaystyle (4,6)} and (2,3){\displaystyle (2,3)}, then the slope would be m=3−62−4{\displaystyle m={\frac {3-6}{2-4}}} m=3−62−4{\displaystyle m={\frac {3-6}{2-4}}} simplifies to −3−2{\displaystyle {\frac {-3}{-2}}} which is equal to 32{\displaystyle {\frac {3}{2}}}. The slope of the line is 32x{\displaystyle {\frac {3}{2}}x}
Using the points (4,6){\displaystyle (4,6)}, the equation would be y−6=32∗(x−4){\displaystyle {y}-{6}={\frac {3}{2}}*({x}-4)}.
To simplify y−6=32∗(x−4){\displaystyle {y}-{6}={\frac {3}{2}}*({x}-4)}, first multiply all of the numbers in parentheses by the outer value to get y−6=32x−6{\displaystyle {y}-{6}={\frac {3}{2}}{x}-6} Isolate the y{\displaystyle y} on one side of the equation by adding 6{\displaystyle 6} to both sides to get y=32x+0{\displaystyle {y}={\frac {3}{2}}{x}+0}. This is the equation of your first line.
The opposite reciprocal of 32x{\displaystyle {\frac {3}{2}}x} is −23x{\displaystyle {-}{\frac {2}{3}}x}.
Using the coordinates from the perpendicular line (6,1){\displaystyle (6,1)} fill in the equation: y−1=−23(x−6){\displaystyle {y}-{1}={-}{\frac {2}{3}}({x}-{6})}. Simplify the equation so that it reads y−1=−23x−6{\displaystyle y-{1}={-}{\frac {2}{3}}x-6} Isolate the y{\displaystyle y} by adding 1{\displaystyle 1} to both sides. The equation now reads y=−23x−5{\displaystyle y={-}{\frac {2}{3}}x-5}. This is the final equation for the perpendicular line.
