8E - Parallel and perpendicular lines

8E - Parallel and perpendicular lines

Matching directions and right angles can be expressed using gradients. You will find parallel and perpendicular lines, handle horizontal and vertical directions, and combine these ideas with midpoints and intersections to solve exact coordinate problems.

Parallel lines

Can you find the line through (1,4)(1,4) that never meets 2x−3y=62x-3y=6?

Rearranging the given equation gives y=23x−2y=\tfrac23x-2. A second line with the same rise for every horizontal step has the same direction. If it has a different intercept, the vertical separation stays constant, so the lines never meet.

Parallel lines

Parallel lines are distinct lines in the same plane that never meet. Nonvertical parallel lines have equal gradients and different yy-intercepts. Two distinct vertical lines are also parallel.

If both gradient and intercept agree, the equations describe one coincident line. For example, 4x−6y=124x-6y=12 is just 2x−3y=62x-3y=6 multiplied by 22; these are two equations of the same line.

Worked example 1: constructing a parallel line

Question. Find the equation of the line through (1,4)(1,4) parallel to 2x−3y=62x-3y=6.

Plan. Extract the given line's gradient and use it with the new point.

y=23x−2rearrange the given liney−4=23(x−1)retain its gradienty=23x+103isolate y\begin{aligned} y&=\frac23x-2 &&\text{rearrange the given line}\\ y-4&=\frac23(x-1) &&\text{retain its gradient}\\ y&=\frac23x+\frac{10}{3} &&\text{isolate }y \end{aligned}

Answer. y=23x+103y=\tfrac23x+\tfrac{10}{3}, or 2x−3y=−102x-3y=-10.

Check. (1,4)(1,4) satisfies the new equation. Its intercept 103\tfrac{10}{3} differs from −2-2, so the equal-gradient lines are distinct.

Now make the gradient depend on a parameter. Equal slopes determine the parameter; the intercepts still decide whether the resulting lines are distinct.

Worked example 2: a parallel-line parameter

Question. Find kk so that kx−3y=12kx-3y=12 is parallel to 2x−3y=62x-3y=6.

Plan. Compare gradients after putting both equations into y=mx+cy=mx+c form.

kx−3y=12⇒y=k3x−42x−3y=6⇒y=23x−2k3=23⇒k=2\begin{aligned} kx-3y=12&\quad\Rightarrow\quad y=\frac{k}{3}x-4\\ 2x-3y=6&\quad\Rightarrow\quad y=\frac23x-2\\ \frac{k}{3}&=\frac23\quad\Rightarrow\quad k=2 \end{aligned}

Answer. k=2k=2.

Check. The resulting lines have equations y=23x−4y=\tfrac23x-4 and y=23x−2y=\tfrac23x-2. Their vertical separation is always 22, so they cannot meet.

For vertical lines there is no finite gradient to compare: x=1x=1 and x=4x=4 are distinct and parallel because their horizontal separation stays 33.

Watch out

Compare gradients after accounting for the coefficient of yy. In 2x−3y=62x-3y=6, the gradient is 23\tfrac23, not 22 or −23-\tfrac23.

Use the same construction with a falling line.

Perpendicular directions

A right-angle turn exchanges the roles of horizontal and vertical changes and reverses one direction.

For example, a direction that moves 33 across and 22 up has gradient 23\tfrac23. Turning that direction through a right angle gives a move of 22 left and 33 up, with gradient −32-\tfrac32. Their product is −1-1.

More generally, rotating a right triangle with horizontal change aa and vertical change bb gives changes −b-b and aa. When both are nonzero, the two gradients are b/ab/a and −a/b-a/b.

Perpendicular gradients

Two nonvertical lines with finite gradients m1m_1 and m2m_2 are perpendicular exactly when

m1m2=−1.m_1m_2=-1.

Thus, for a nonzero gradient mm, the perpendicular gradient is its negative reciprocal:

m⊥=−1m.m_{\perp}=-\frac1m.

A horizontal line and a vertical line are perpendicular; treat them separately.

Not on the formulae sheet: learn it.

Taking the negative reciprocal means both reversing the fraction and changing its sign. For example, the negative reciprocal of −32-\tfrac32 is 23\tfrac23.

Worked example 3: a perpendicular line through a point

Question. Find the equation through (1,−2)(1,-2) perpendicular to 3x+2y=83x+2y=8.

Plan. Find the original gradient, take its negative reciprocal and use point-gradient form.

y=−32x+4original gradient −32m⊥=23negative reciprocaly+2=23(x−1)through (1,−2)y=23x−83simplify\begin{aligned} y&=-\frac32x+4 &&\text{original gradient }-\frac32\\ m_{\perp}&=\frac23 &&\text{negative reciprocal}\\ y+2&=\frac23(x-1) &&\text{through }(1,-2)\\ y&=\frac23x-\frac83 &&\text{simplify} \end{aligned}

Answer. y=23x−83y=\tfrac23x-\tfrac83.

Check. Substitution of x=1x=1 gives y=−2y=-2, and (−32)(23)=−1(-\tfrac32)(\tfrac23)=-1, confirming the point and right angle independently.

Keep the required point but replace the original line with a vertical line. Its gradient is undefined, so taking a reciprocal is not available.

Worked example 4: perpendicular to a vertical line

Question. Find the equation through (1,−2)(1,-2) perpendicular to x=5x=5.

Plan. A perpendicular to a vertical line is horizontal, so fix the point's yy-coordinate.

y=−2.y=-2.

Answer. y=−2y=-2.

Check. It contains (1,−2)(1,-2) and meets x=5x=5 at (5,−2)(5,-2). One direction is horizontal and the other vertical, so their angle is a right angle.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account