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E3.7 Perpendicular lines

Syllabus
0580–2028–2029
Topic
E3.7
Level
Extended

Find perpendicular lines and bisectors

Two non-vertical, non-horizontal perpendicular lines meet at a right angle and have gradients that are negative reciprocals. This reverses the rise/run ratio and changes its sign.

m1m2=−1⟺m2=−1m1m_1m_2=-1\qquad\Longleftrightarrow\qquad m_2=-\frac{1}{m_1}

First expose the given gradient. From 2y=3x+12y=3x+1, obtain y=(3/2)x+1/2y=(3/2)x+1/2, so m1=3/2m_1=3/2. The perpendicular gradient is m2=−2/3m_2=-2/3.

For a perpendicular line through a specified point, write y=m2x+cy=m_2x+c, substitute the point to find cc, then simplify and check. For example, perpendicular to y=−frac12x+7y=- frac12x+7 through (3,5)(3,5) gives m2=2m_2=2 and 5=2(3)+c5=2(3)+c, so y=2x−1y=2x-1.

A perpendicular bisector must be both perpendicular to the segment and pass through its midpoint. Find the segment gradient, take its negative reciprocal, calculate the midpoint, then substitute that midpoint into the new line equation.

A(−3,8), B(9,−2):M=(3,3),mAB=−56,m⊥=65,y=65x−35A(-3,8),\ B(9,-2):\quad M=(3,3),\quad m_{AB}=-\frac56,\quad m_{\perp}=\frac65,\quad y=\frac65x-\frac35

The final line has gradient 6/56/5 and contains (3,3)(3,3) because (6/5)(3)−3/5=3(6/5)(3)-3/5=3. Its gradient product with −5/6-5/6 is −1-1, confirming perpendicularity.

Do not only change the sign: the reciprocal is also required. Horizontal and vertical lines are the special pair—y=ky=k is perpendicular to x=ax=a—so the product rule is not used when a gradient is undefined.