E4.6 - Perpendicular lines and bisectors (Extended)

E4.6 - Perpendicular lines and bisectors (Extended)

A line rises 2 units for every 3 units to the right. What direction would cross it at a right angle? E4.6 develops perpendicular gradients and equations, then combines them with a midpoint to find a perpendicular bisector.

Perpendicular gradient

Turn a right-angled gradient triangle through 90∘90^\circ. A movement 3 right and 2 up becomes 2 left and 3 up. The first direction has gradient 2/32/3; the turned direction has gradient 3/(−2)=−3/23/(-2)=-3/2.

[DIAGRAM: asset_name: Turning a gradient direction through a right angle; asset_slug: perpendicular_gradient_turn; caption: A quarter-turn exchanges horizontal and vertical changes and reverses one sign.; description: Equal-scale coordinate axes show two lines through the origin with gradients two thirds and minus three halves. A blue direction triangle has run three and rise two, and a sage triangle has run minus two and rise three. A right-angle marker at the origin confirms the perpendicular directions.]
A quarter-turn exchanges horizontal and vertical changes and reverses one sign.

Perpendicular gradients

For two non-vertical lines with non-zero gradients,

m1m2=−1,m2=−1m1.m_1m_2=-1,\qquad m_2=-\frac1{m_1}.

The second gradient is the negative reciprocal of the first. A horizontal line and a vertical line are perpendicular; their gradients do not fit the product rule because the vertical gradient is undefined.

Not in the supplied formula list: learn it.

The quarter-turn explains both parts of the rule: exchange numerator and denominator, then change the sign.

Worked example 1: a positive fractional gradient

Question. Find the gradient of a line perpendicular to y=34x+2y=\frac34x+2.

Plan. Take the negative reciprocal of 3/43/4.

m=34m⊥=−13/4=−43\begin{aligned} m&=\frac34\\ m_{\perp}&=-\frac1{3/4}=-\frac43 \end{aligned}

Answer. −4/3-4/3.

Check. (3/4)(−4/3)=−1(3/4)(-4/3)=-1, as required.

Now the original gradient is negative. Its negative reciprocal is positive.

Worked example 2: first revealing the gradient

Question. Find the gradient of a line perpendicular to 3y=−4x+63y=-4x+6.

Plan. Divide the whole equation by 3, then take the negative reciprocal.

y=−43x+2divide every term by 3m⊥=−1−4/3=34\begin{aligned} y&=-\frac43x+2 &&\text{divide every term by 3}\\ m_{\perp}&=-\frac1{-4/3}=\frac34 \end{aligned}

Answer. 3/43/4.

Check. (−4/3)(3/4)=−1(-4/3)(3/4)=-1. A negative slope and a positive slope can cross at a right angle.

For y=5y=5, the gradient is zero. A perpendicular is vertical, such as x=2x=2; do not attempt to evaluate −1/0-1/0. Conversely, every perpendicular to x=−4x=-4 is horizontal and has gradient zero.

Watch out

Changing only the sign is insufficient: a perpendicular to gradient 5 has gradient −1/5-1/5, not −5-5. Taking only the reciprocal is also insufficient.

Separate the ordinary case from the vertical case.

Equation through a point

The perpendicular gradient sets the direction. Substituting a specified point into y=mx+cy=mx+c locates the required perpendicular line.

Worked example 3: a positive original gradient

Question. Find the equation of the line perpendicular to y=2x−7y=2x-7 through (4,1)(4,1).

Plan. Find the negative reciprocal of 2, then use the point to find cc.

m⊥=−121=−12(4)+cuse the pointc=3\begin{aligned} m_{\perp}&=-\frac12\\ 1&=-\frac12(4)+c &&\text{use the point}\\ c&=3 \end{aligned}

Answer. y=−12x+3y=-\frac12x+3.

Check. 2(−1/2)=−12(-1/2)=-1, and at x=4x=4 the proposed line gives −2+3=1-2+3=1. Both the direction and the point are correct.

Change only the sign of the original gradient next. The perpendicular direction reverses its sign too.

Worked example 4: a negative original gradient

Question. Find the equation of the line perpendicular to y=−2x−7y=-2x-7 through (4,1)(4,1).

Plan. Use perpendicular gradient 1/21/2, then substitute the point.

y=12x+c1=12(4)+cc=−1\begin{aligned} y&=\frac12x+c\\ 1&=\frac12(4)+c\\ c&=-1 \end{aligned}

Answer. y=12x−1y=\frac12x-1.

Check. (−2)(1/2)=−1(-2)(1/2)=-1 and (1/2)(4)−1=1(1/2)(4)-1=1.

A perpendicular to the horizontal line y=−3y=-3 through (5,2)(5,2) is x=5x=5. A perpendicular to the vertical line x=−3x=-3 through that point is y=2y=2. In these cases the relevant coordinate gives the equation immediately.

Exam tip

Show the perpendicular gradient and the point substitution. They are separate decisions: a correct gradient alone does not locate the line.

Simplify the given equation before finding its perpendicular.

Continue this lesson.

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