45 min readNo calculatorNeeds: quadratic functions and equations
Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — distance, midpoint and perpendicular bisector
The points are A(−1,4) and B(5,−2). Find the length of AB, the midpoint of AB, and the equation of the perpendicular bisector of AB.
1Differences: Δx=5−(−1)=6 and Δy=−2−4=−6Take them in the same order both times, or the gradient sign will flip.
2∣AB∣=62+(−6)2=72=62Square first, then add, then take the root.
3Midpoint =(2−1+5,24+(−2))=(2,1)The midpoint is just the average of each coordinate.
4Gradient of AB=6−6=−1, so the perpendicular gradient is 1Negative reciprocal: flip it and change the sign. The reciprocal of −1 is −1, and changing the sign gives +1.
5y−1=1(x−2), so y=x−1The bisector must pass through the midpoint, which is why we found it first.
Example 2 — reading a circle out of its general form
Find the centre and radius of the circle x2+y2−6x+4y−12=0.
1Compare with x2+y2+2gx+2fy+c=0The coefficients of x and y are 2g and 2f, so they must be halved.
22g=−6⇒g=−3; 2f=4⇒f=2; c=−12Keep the signs exactly as they appear.
3Centre =(−g,−f)=(3,−2)Halve and change the sign — two operations, both needed.
4r=g2+f2−c=9+4−(−12)c is negative, so −c is positive. This sign is the usual slip.
5=25=5Centre (3,−2), radius 5.
Example 3 — the tangent at a point on a circle
Find the equation of the tangent to the circle x2+y2=25 at the point P(3,4).
1Check P is on the circle: 32+42=25 ✓Worth ten seconds. If P is not on the circle the whole method changes.
2The centre is the origin, so the gradient of the radius OP is 34For x2+y2=r2 the centre is always (0,0).
3The tangent is perpendicular to the radius, so its gradient is −43Negative reciprocal of 34.
4y−4=−43(x−3)Use the point of contact, not the centre.
54y−16=−3x+9, so 3x+4y=25Check: 3(3)+4(4)=9+16=25 ✓