CG5 - Gradient, intercept, tangent and normal

CG5 - Gradient, intercept, tangent and normal

A point fixes a location; a gradient fixes a direction. Combine them to find lines, perpendicular bisectors, and tangents and normals to circles and curves. You will use coordinate geometry and repeated intersections to justify the answers, keeping every line tied to the point it must pass through.

Gradients, intercepts and line equations

A circle has centre (1,−2)(1, -2) and radius 55. At (4,2)(4, 2), which line touches the circle without cutting across it, and which line points straight towards its centre? The centre-to-point direction will decide both answers; first we need a precise way to describe a direction.

You already know

A straight line has equation y=mx+cy=mx+c unless it is vertical. Its vertical intercept is (0,c)(0, c); a horizontal intercept is found by putting y=0y=0. You can solve two line equations simultaneously to find their intersection.

From (1,2)(1, 2) to (4,8)(4, 8), a horizontal increase of 33 accompanies a vertical increase of 66. The vertical change per unit horizontal change is 6/3=26/3=2. This ratio, rather than the length of the segment, is its gradient.

Gradient

The gradient of a non-vertical straight line is the change in yy divided by the change in xx. Positive gradients rise from left to right, negative gradients fall, and a horizontal line has gradient zero. A vertical line has no finite gradient because its horizontal change is zero.

Any two distinct points on the same non-vertical line give the same ratio. In particular, a line through (x1,y1)(x_1, y_1) with gradient mm satisfies y−y1=m(x−x1)y-y_1=m(x-x_1).

Gradient and point-gradient form

m=y2−y1x2−x1,y−y1=m(x−x1)m=\frac{y_2-y_1}{x_2-x_1}, \qquad y-y_1=m(x-x_1)

For x2≠x1x_2\ne x_1. In y=mx+cy=mx+c, cc is the vertical intercept value. A vertical line through (a,b)(a, b) is x=ax=a, and a horizontal line through it is y=by=b.
Not on the formulae sheet: learn it.

Use the gradient to set the direction, then the point to place the line. A gradient on its own describes infinitely many parallel lines.

Worked example 1: a line through two points

Question. Find the equation of the line through A(−1,5)A(-1, 5) and B(3,−3)B(3, -3).

Plan. Calculate the gradient, then use point-gradient form through AA.

m=−3−53−(−1)=−2matching coordinate changesy−5=−2(x+1)line through Ay=−2x+3expand and rearrange\begin{aligned} m&=\frac{-3-5}{3-(-1)}=-2 && \text{matching coordinate changes}\\ y-5&=-2(x+1) && \text{line through }A\\ y&=-2x+3 && \text{expand and rearrange} \end{aligned}

Why x+1x+1? The expression is x−(−1)x-(-1), so it vanishes at the chosen point.

Answer. y=−2x+3y=-2x+3.

Check. At x=−1x=-1, y=5y=5; at x=3x=3, y=−3y=-3. The vertical intercept is (0,3)(0, 3).

Change only the vertical coordinate of BB from −3-3 to 77. Recalculate the gradient; the old direction no longer applies.

Worked example 2: a fractional gradient

Question. Find the equation of the line through A(−1,5)A(-1, 5) and B(3,7)B(3, 7).

Plan. Form the new gradient and place the line through the unchanged point AA.

m=7−53−(−1)=12rise 2 over run 4y−5=12(x+1)point-gradient form2y=x+11clear the fraction\begin{aligned} m&=\frac{7-5}{3-(-1)}=\frac12 && \text{rise }2\text{ over run }4\\ y-5&=\frac12(x+1) && \text{point-gradient form}\\ 2y&=x+11 && \text{clear the fraction} \end{aligned}

Why leave the fraction exact? Rounding a gradient can move the line away from the given points.

Answer. 2y=x+112y=x+11, equivalently y=12x+112y=\frac12x+\frac{11}{2}.

Check. 2(5)=−1+112(5)=-1+11 and 2(7)=3+112(7)=3+11, so both points satisfy it.

The horizontal intercept of the second line is (−11,0)(-11, 0); its vertical intercept is (0,112)(0, \frac{11}{2}). The coefficient of xx is only the gradient after yy has been isolated.

Watch out

Wrong: 2y=x+112y=x+11 has gradient 11. Right: divide by 22 to obtain y=12x+112y=\frac12x+\frac{11}{2}, so its gradient is 12\frac12.

Change the second point's height once more.

Parallel lines and perpendicular bisectors

Parallel non-vertical lines share a gradient; perpendicular non-vertical lines have gradients whose product is −1-1.

To see the second rule, take a direction with horizontal and vertical changes (p,q)(p, q). Rotating it through a right angle gives changes (−q,p)(-q, p). Their gradients are q/pq/p and −p/q-p/q, whose product is −1-1, when both are finite and non-zero. Horizontal and vertical lines are the separate perpendicular pair.

Parallel and perpendicular directions

m2=m1for parallel lines,m2=−1m1for perpendicular linesm_2=m_1\quad\text{for parallel lines}, \qquad m_2=-\frac1{m_1}\quad\text{for perpendicular lines}

The perpendicular formula requires a finite non-zero m1m_1. A line perpendicular to a horizontal line is vertical, and a line perpendicular to a vertical line is horizontal.
Not on the formulae sheet: learn it.

A line parallel to y=2x−7y=2x-7, for example, has gradient 22 but can have a different intercept. A perpendicular bisector needs both a new direction and a particular point.

Perpendicular bisector

The perpendicular bisector of a segment passes through its mid-point at right angles to the segment. Every point on it is equidistant from the two endpoints; conversely, every equidistant point lies on it. A line through the mid-point with the wrong gradient bisects the segment but is not its perpendicular bisector.

The equidistance claim comes from two right-angled triangles sharing the line to the mid-point and having equal half-segments. Pythagoras gives equal distances to the endpoints. This also explains why the whole line, not just the mid-point, meets the equal-distance condition.

Worked example 3: a perpendicular bisector

Question. Find the equation of the perpendicular bisector of ABAB, where A(−2,1)A(-2, 1) and B(4,5)B(4, 5).

Plan. Find the mid-point and segment gradient, take the perpendicular gradient, then use the mid-point to place the line.

M=(−2+42,1+52)=(1,3)mid-pointmAB=5−14−(−2)=23segment gradientm⊥=−32negative reciprocaly−3=−32(x−1)through M3x+2y=9clear fractions\begin{aligned} M&=\left(\frac{-2+4}{2}, \frac{1+5}{2}\right)=(1, 3) && \text{mid-point}\\ m_{AB}&=\frac{5-1}{4-(-2)}=\frac23 && \text{segment gradient}\\ m_{\perp}&=-\frac32 && \text{negative reciprocal}\\ y-3&=-\frac32(x-1) && \text{through }M\\ 3x+2y&=9 && \text{clear fractions} \end{aligned}

Why use the mid-point rather than an endpoint? The bisector meets the segment halfway along, not at either end.

Answer. 3x+2y=93x+2y=9.

Check. (1,3)(1, 3) satisfies the line, and 23×(−32)=−1\frac23\times(-\frac32)=-1, so it has both required properties.

Change only BB's vertical coordinate from 55 to −3-3. Both the mid-point and the perpendicular direction must be recalculated.

Worked example 4: a falling segment

Question. Find the perpendicular bisector of A(−2,1)A(-2, 1) and B(4,−3)B(4, -3).

Plan. Repeat the mid-point and perpendicular-gradient method with the new endpoint.

M=(1,−1)average the coordinatesmAB=−3−14−(−2)=−23segment gradientm⊥=32negative reciprocaly+1=32(x−1)through the mid-point3x−2y=5rearrange\begin{aligned} M&=(1, -1) && \text{average the coordinates}\\ m_{AB}&=\frac{-3-1}{4-(-2)}=-\frac23 && \text{segment gradient}\\ m_{\perp}&=\frac32 && \text{negative reciprocal}\\ y+1&=\frac32(x-1) && \text{through the mid-point}\\ 3x-2y&=5 && \text{rearrange} \end{aligned}

Why is the perpendicular gradient positive? A negative gradient multiplied by its perpendicular gradient must give −1-1.

Answer. 3x−2y=53x-2y=5.

Check. The point (3,2)(3, 2) is on this line. Its squared distances from the endpoints are 52+12=265^2+1^2=26 and (−1)2+52=26(-1)^2+5^2=26, equal as expected.

For the special case of a horizontal segment, skip the reciprocal calculation and write a vertical line through the mid-point.

Watch out

Wrong: the perpendicular gradient to 23\frac23 is −23-\frac23. Changing only the sign does not give a right angle. Right: change the sign and take the reciprocal, giving −32-\frac32.

Now make the segment horizontal by changing just the second height.

Tangents and normals to circles

At a point on a circle, the radius is perpendicular to the tangent, so its gradient supplies exactly the direction we need.

Tangent and normal

A tangent gives a smooth curve's direction at the chosen point. For a circle, it meets the circle only at that point and is perpendicular to its radius there. The normal is the line through the point perpendicular to the tangent; for a circle it is the line through the centre and that point.

There is a short reason the perpendicular radius works. Let CC be the centre and PP the point on the circle. If Q≠PQ\ne P is anywhere on the perpendicular line through PP, then CQ2=CP2+PQ2>r2CQ^2=CP^2+PQ^2>r^2. Every other point on that line is outside the circle. It touches at PP alone. This tangent-radius result is not on the formulae sheet: learn it.

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