E3.7 - Perpendicular lines and bisectors

E3.7 - Perpendicular lines and bisectors

A right angle has an algebraic signature: perpendicular sloping lines have related gradients. You will explain that relationship, find a perpendicular line through a point and combine it with midpoint calculations to find perpendicular bisectors, as required by E3.7.

Perpendicular gradients

If a line has gradient 23\frac23, why is a perpendicular line's gradient −32-\frac32, rather than −23-\frac23? Draw a direction triangle with a run of 33 right and a rise of 22 up. Turn the whole triangle anticlockwise through 90∘90^\circ: the three-unit horizontal side points up, and the two-unit vertical side points left. The new sloping side has signed run −2-2 and rise 33, giving gradient 3/(−2)=−3/23/(-2)=-3/2.

[DIAGRAM: asset_name: A quarter-turn exchanges rise and run; asset_slug: perpendicular_gradient_turn; caption: Turning the whole right triangle through a right angle exchanges the changes and reverses one sign.; description: Equal-scale axes show O at the origin, A at three comma two and B at negative two comma three. OA has horizontal change three and vertical change two. Its anticlockwise quarter-turned copy OB has horizontal change negative two and vertical change three. Both right-angled direction triangles are shown, and a right-angle marker between OA and OB identifies the perpendicular directions.]
Turning the whole right triangle through a right angle exchanges the changes and reverses one sign.

This works with signed changes too. Call the original non-zero horizontal change aa and its non-zero vertical change bb. The gradient is b/ab/a. A quarter-turn gives horizontal change −b-b and vertical change aa, so the new gradient is −a/b-a/b. If bb is negative, the original line falls and the perpendicular line rises.

Perpendicular gradients

Two non-vertical, non-horizontal straight lines are perpendicular exactly when

m2=−1m1.m_2=-\frac1{m_1}.

Equivalently,

m1m2=−1.m_1m_2=-1.

Their gradients are negative reciprocals: invert the fraction and change its sign. Neither gradient is zero. A horizontal line is perpendicular to a vertical line; handle that pair separately. Not given in the List of formulas: learn it.

The rule works in both directions. Turning a line through 90∘90^\circ produces a line with the calculated gradient; every line with that gradient has the same direction. Thus the rule can also establish that two given sloping lines are perpendicular.

Worked example 1: perpendicular to a rising line

Question. Find the gradient of a line perpendicular to y=23x+1y=\frac23x+1.

Plan. Read the gradient, invert its fraction and change the sign.

m1=23coefficient of xm2=−12/3=−32take the negative reciprocal\begin{aligned} m_1&=\frac23 && \text{coefficient of }x\\ m_2&=-\frac1{2/3}=-\frac32 && \text{take the negative reciprocal} \end{aligned}

Answer. −32-\frac32.

Check. The original direction goes three right and two up. A right-angle turn gives two left and three up, whose gradient is −3/2-3/2, agreeing with the calculation.

Change only the original gradient's sign. Its perpendicular gradient must change sign as well.

Worked example 2: perpendicular to a falling line

Question. Find the gradient of a line perpendicular to y=−23x+1y=-\frac23x+1.

Plan. Keep the original minus sign when dividing into negative one.

m1=−23read the signed coefficientm2=−1−2/3=32two negative signs cancel\begin{aligned} m_1&=-\frac23 && \text{read the signed coefficient}\\ m_2&=-\frac1{-2/3}=\frac32 && \text{two negative signs cancel} \end{aligned}

Answer. 32\frac32.

Check. A direction of three right and two down becomes two right and three up after an anticlockwise quarter-turn. Its gradient is 3/23/2, as found.

For a horizontal line, the rise is zero. Turning its direction through a right angle produces a vertical line, whose run is zero and gradient undefined. Do not try to calculate −1/0-1/0. Conversely, a perpendicular to a vertical line is horizontal, with gradient zero.

Watch out

Changing only the sign is insufficient: 23×(−23)=−49\frac23\times(-\frac23)=-\frac49, not −1-1. Use −32-\frac32. A special case is gradient 11, whose negative reciprocal happens to be −1-1.

The opening reciprocal appears because a quarter-turn swaps the horizontal and vertical changes, as well as reversing one direction.

A perpendicular line through a point

A perpendicular gradient fixes direction, but not position. To find the required line, first put the reference equation into y=mx+cy=mx+c, take the negative reciprocal of its gradient, then substitute the required point to find the new intercept.

You already know

For a line with known gradient pp, write y=px+cy=px+c. A point (u,v)(u,v) on it gives v=pu+cv=pu+c, so c=v−puc=v-pu. This is the same point-substitution method used for parallel lines; only the gradient choice changes.

The point is on the new line; it need not be on the reference line. There is exactly one perpendicular line through a specified point.

Worked example 3: choosing direction and position

Question. Find the line perpendicular to 4y=2x+84y=2x+8 and passing through (3,1)(3,1). Give its equation in the form y=mx+cy=mx+c.

Plan. Rearrange to read the reference gradient, find the perpendicular gradient, then use the point.

y=12x+2divide every term by fourp=−11/2=−2perpendicular gradient1=−2(3)+csubstitute into y=−2x+cc=7solve for the intercept\begin{aligned} y&=\frac12x+2 && \text{divide every term by four}\\ p&=-\frac1{1/2}=-2 && \text{perpendicular gradient}\\ 1&=-2(3)+c && \text{substitute into }y=-2x+c\\ c&=7 && \text{solve for the intercept} \end{aligned}

Answer. y=−2x+7y=-2x+7.

Check. At x=3x=3, the new equation gives y=1y=1. Its direction is one right and two down; reversing that direction gives one left and two up, a quarter-turn of the reference direction two right and one up. Both the point and the right-angle condition are satisfied.

Change only the point's second coordinate from 11 to −1-1. The required direction stays fixed; the intercept must change.

Worked example 4: moving the point

Question. Find the line perpendicular to 4y=2x+84y=2x+8 and passing through (3,−1)(3,-1), in the form y=mx+cy=mx+c.

Plan. Keep the perpendicular gradient −2-2, then substitute the new point.

y=−2x+csame perpendicular direction−1=−2(3)+cuse the new heightc=5add six to both sides\begin{aligned} y&=-2x+c && \text{same perpendicular direction}\\ -1&=-2(3)+c && \text{use the new height}\\ c&=5 && \text{add six to both sides} \end{aligned}

Answer. y=−2x+5y=-2x+5.

Check. At x=3x=3, it gives −6+5=−1-6+5=-1. The new line is two units below the previous perpendicular line at every input, matching the point's two-unit fall; its unchanged direction is still perpendicular to the reference line.

For an axis-parallel reference line, choose the other direction directly. A perpendicular to x=−4x=-4 through (2,5)(2,5) is horizontal, so y=5y=5. A perpendicular to y=−4y=-4 through the same point is vertical, so x=2x=2. These cases need no reciprocal or intercept calculation.

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