Coordinate Geometry: Straight Lines and Circles

1Distance and section

The distance between (x_1,y_1) and (x_2,y_2) is \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}. A point dividing the same segment in the ratio m:n (internally) is \left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right) — weighted averages with the nearer endpoint carrying the larger weight.

Figure. P sits one-third of the way from A to B when the ratio is 1:2 — closer to the endpoint whose weight is larger in the complementary sense.

Reading the segment

1. Distance firstSquare the differences of the coordinates, add, and take the positive square root — distance is never signed.
2. Ratio nextFor a point dividing A:B in m:n, weight B's coordinates by m and A's by n, then divide by m+n.
3. Check the nearer endIf m>n the section point sits closer to B; swapping m and n swaps which end it hugs.

2Line forms and the angle between them

Three writings of the same line: slope-intercept y=mx+c, point-slope y-y_1=m(x-x_1), and general ax+by+c=0 with slope -a/b when b\neq 0. The angle between lines of slopes m_1 and m_2 is \tan\theta=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|; they are parallel when m_1=m_2 and perpendicular when m_1m_2=-1. The product test decides perpendicular — do not reach for the tangent formula once the slopes already multiply to minus one.

Figure. Both lines share the origin. The steeper series is slope 2; the shallow series is slope 1/2 — the pair used in the worked tan θ.

From form to angle

1. Read the slopeFrom ax+by+c=0, m=-a/b. From y=mx+c, the slope is already m.
2. Plug into the tangent formulaCompute \tan\theta=|(m_1-m_2)/(1+m_1m_2)|. If the denominator is zero the lines are perpendicular.
3. Name the special casesEqual slopes mean parallel; product -1 means perpendicular — no need to evaluate \theta.

3Distance from a point to a line

The perpendicular distance from (x_1,y_1) to ax+by+c=0 is \dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}. Between parallel lines ax+by+c_1=0 and ax+by+c_2=0 the gap is \dfrac{|c_1-c_2|}{\sqrt{a^2+b^2}} — same denominator, difference of the constants only.

Figure. The dashed segment is the perpendicular from the point to the line — its length is the formula, not the slant along the axes.

Perpendicular distance

1. Plug the point into SEvaluate |ax_1+by_1+c| — the absolute value keeps distance positive on either side of the line.
2. Divide by the normDivide by \sqrt{a^2+b^2}, the length of the normal vector (a,b).
3. Parallel linesWhen the a,b coefficients match, only |c_1-c_2| changes in the numerator.

4Perpendicular to ax+by+c=0

Perpendicular lines have slopes whose product is -1. For a line ax+by+c=0 any perpendicular has the form bx-ay+k=0 — swap the coefficients, flip one sign, and leave the constant free to fix the particular line through a given point.

Figure. Slopes −3/4 and 4/3 multiply to −1 at the marked meeting point. The free constant k only slides the second line until it passes through the point you need.

Writing a perpendicular

1. Start from ax+by+c=0Read a and b. The perpendicular family is bx-ay+k=0.
2. Fix k if neededPass the line through a given point (x_0,y_0) to solve for k=ay_0-bx_0.
3. Check the slopesOriginal slope -a/b; new slope b/a; product -1 when both are defined.

5Foot of the perpendicular

The foot of the perpendicular from (x_1,y_1) to ax+by+c=0, and the image of the point in the line, both come from one parametric step: \dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=-\dfrac{ax_1+by_1+c}{a^2+b^2}. The shared value of that parameter is the signed travel along the normal; the foot is one step, the image is two.

Figure. One dashed normal from P through F to the image P′. F is the midpoint of P and P′ — that is what “the image is two steps” means on the page.

Foot, then image

1. Form the parameterCompute t=-(ax_1+by_1+c)/(a^2+b^2). The same t feeds both coordinates.
2. Step once for the footx=x_1+at, y=y_1+bt is the foot on the line.
3. Step twice for the imageThe reflection is (x_1+2at,\,y_1+2bt), equivalently 2\cdot\mathrm{foot}-(x_1,y_1).

6Centre and radius from the general circle

The standard circle is (x-h)^2+(y-k)^2=r^2. Expanded, the general form x^2+y^2+2gx+2fy+c=0 has centre (-g,-f) and radius \sqrt{g^2+f^2-c} — provided g^2+f^2-c>0. Read g and f as half the coefficients of x and y; the centre is their negatives, and that sign flip is the usual miss.

Figure. Standard form (x-h)^2+(y-k)^2=r^2 is a round locus about the centre; here h=k=0, r=6. The general-form centre (-g,-f) is the same reading after completing the square.

Reading the general circle

1. Halve the linear termsCompare with x^2+y^2+2gx+2fy+c=0: g is half the x-coefficient, f half the y-coefficient.
2. Centre is the negativesCentre =(-g,-f). Sign-flipping here is the standard exam trap.
3. Radius from g,f,cr=\sqrt{g^2+f^2-c}. If the quantity under the root is negative there is no real circle.

7Point position via S_1

Write the circle as S\equiv x^2+y^2+2gx+2fy+c=0. To test a point (x_1,y_1), evaluate S_1=x_1^2+y_1^2+2gx_1+2fy_1+c: positive means outside, zero on the circle, negative inside. The same S_1 later becomes the square of the tangent length from an external point.

Figure. For S=x^2+y^2-25, a test point with S_1<0 lies inside the circle and one with S_1>0 lies outside — the sign of S_1 is the position test.

8Tangents and tangent length

A line is tangent to a circle when its distance from the centre equals the radius. The tangent to x^2+y^2=r^2 at (x_1,y_1) on the circle is xx_1+yy_1=r^2. From an external point the length of either tangent is \sqrt{S_1} — the same S_1 that tested position. Equal tangent lengths are a theorem, not a second computation.

Tangent length from a point

1. Confirm outsideEvaluate S_1. If S_1\le 0 there is no real tangent pair from that point.
2. Take the rootLength =\sqrt{S_1}=\sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c}.
3. At a point on x^2+y^2=r^2The contact chord / tangent is xx_1+yy_1=r^2, not the polar of an outside point.