Coordinate Geometry Examples

    You should get hold of some precalculus textbook, such as Schaum Series. In other countries students pay for their math textbooks during high school, so they keep them when going to college. But in America college students do not know their math because they no longer possess their textbooks in high school. So you must spend money judiciously to acquire good reference texts, on algebra, geometry, and trig.
    Formulas. You should memorise the basic 4 or 5 equations of a straight line, the formulas for distance between points, the distance of a point from a line, the formula of the tangent at a point on a circle or ellipse or parabola. You should memorise them much like memorising the basic 30 formulas of trigonometry.
    Of the following examples you must learn the logic, the techniques, and you must endeavour to do the proofs again, by yourself. In tests I shall offer you the same or similar problems, but never with the same numerics. So you must learn how to do it. Notice in reducing
(x - 2)² = (y + 1)² - 4 to x² - y² - 4 x - 2 y + 7 = 0, I used mental calculation to gather the x² and y² terms, the x and y terms, and the constant terms all onto one side, totally without writing any more lines. This brevity contributes to the elegance of a proof, and you must learn to do such mental arithmetic. In high school, it is a no-no not to write out all details, but from now on, it is a beauty and an art to be brief over such computational details.

    Some smart things: Whenever you mention a point, give it a name as well as its coordinates, such as P_i(x_i, y_i). Whenever you mention a line, such as AB, it is sometimes wise to give it a name such as L₂ and an equation, and call it (e2). Later on you can mention the line as AB or L₂ or (e2), totally interchangeably. You need not say the equation (e2) "represents" the line. You can simply say "the equation (e2) is the line L₂". Similarly, you need not say (x_i, y_i) "are the coords of" or "represents" the point P_i . You can simply say (x_i, y_i) is the point P_i . When I say (a, b) = (x, y), it means a = x and b = y. (a, b) may mean the point with a, b as coords, or a vector from the origin to the point. It doesn't matter.

    Problem 1. Find the perpendicular bisector L between the points A(3, 4) and B( -1, 10).
    Solution 1. The slope of AB is (10 - 4)/( -1 - 3) = 6/ - 4 = -3/2.
Hence the slope of the line L is -1 / ( -3/2) = 2/3.
The midpoint M of AB is ((½)(3 + -1), (½)(4 + 10)) = (1, 7).
Hence the line L is the line thru M with slope 2/3, with equation (called point-and-slope form) being
y - 7 = (2/3) (x -1), i.e.,
3 y - 2 x - 19 = 0. QED
    Solution 2.(More elegant.) L is the locus (i.e., "the set") of points P(x, y) equidistant from A and B, hence
Ö[(x - 3)² + (y - 4)²] = Ö[(x - -1)² + (y - 10)²],
i.e., x² - 6 x + y² - 8 y + 25 = x² + 2 x + y² - 20 y + 101,
i.e., 12 y - 8 x - 76 = 0,
i.e., 3 y - 2 x - 19 = 0. QED

    Problem 2. Find the angle bisectors of the lines L1: 2 y + x = 4 and L2: y = 2 x - 3, without solving for their intersection point.
    Solution. The two bisectors form the locus of points P(x, y) equidistant from the two lines. Hence using the distance-formula of a point from a line, we get
| 2 y + x - 4 | / Ö(2² + 1²) = | y - 2 x + 3 | / Ö(1² + 2²).
\Either 2 y + x - 4 = y - 2 x + 3 or 2 y + x - 4 = - (y - 2 x + 3).
\Either y + 3 x - 7 = 0 or 3 y - x - 1 = 0,
which are the equations of the angle bisectors of L1 and L2. QED

    Problem 3. Derive the two-point form and intercepts form of the straight line.

    Solution. (a) Let two points P₁(x₁, y₁) and P₂(x₂, y₂) be given.
Let P(x, y) be any point on the line P₁P₂.
Let Q and Q₂ be the projections of P and P₂ upon the line thru P₁ parallel to the x-axis.
Then PQ / P₂Q₂ = P₁Q / P₁Q₂ by similar triangles.
\ (y - y₁) / (y₂ - y₁) = (x - x₁) / (x₂ - x₁),
which is the two-point form of a line through two given points P₁ and P₂.
    (b) We want to find the str. line that intercepts the x-axis at (a, 0) and the y-axis at (0, b).
Using the two point form, we get
(y - 0) / (b - 0) = (x - a) / (0 - a),
i.e., y/b = - x/a + 1,
i.e., x/a + y/b = 1, which is the intercepts form.

    Problem 4. Find a str. line L that goes through the point P(1,2) and
the intersection Q of the lines L1: 2 x - y = 3 and L2: x - 3 y + 1 = 0.
    Solution 1. Solve the equations to find Q, then use two-point form to fine the equation of PQ.
    Solution 2. More elegant. Consider the equation of some line
L_m: (2 x - y - 3) + m (x - 3 y + 1) = 0.
It is formed by combining L1 and L2 with an arbitrary ratio 1 : m.
As m varies, it represents a family of str. lines.
Notice that if Q is (x', y'), then 2 x' - y' - 3 = 0 and x' - 3 y' + 1 = 0.
This proves that L_m goes thru Q, for all m. We want to find a member of this family going through P.
\Substitute the coords of P into L_m, we get
(2 × 1 - 2 - 3) + m (1 - 3 × 2 + 1) = 0, giving
- 3 - 4 m = 0 or m = -3/4.
Substitute this value of m into L_m, we get
(2 x - y - 3) - (3/4) (x - 3 y + 1) = 0,
i.e., 5 x - 5 y - 15 = 0,
i.e., y = x - 3. QED
    Note. I could modify the problem so that the line L instead of having to go through the point P, would have to be perpendicular to another given line, or would be at some specific distance from a given point, or be equidistant from two given points, etc. The same L_m technique would work beautifully.

    Problem 5. Find the orthocenter of the triangle whose sides are
L₁: 2 y - x - 3 = 0, L₂: x + y - 6 = 0, L₃: x = 1.
You are forbidden to solve for the coordinates of the 3 vertices.
    Solution. Let P₃ = L₁ Ç L₂; P₂ = L₃ Ç L₁;
and P₁ = L₂ Ç L₃, be the 3 vertices of the triangle. The family of lines through P₁ is (L₂) + m (L₃) = 0,
or, (x + y - 6) + m (x - 1) = 0.
We want to choose the member that is an altitude thru P₁, i.e., perp. to L₁.
L₁ is y = (½)(x + 3) whose slope is 1/2.
The slope of (x + y - 6) + m (x - 1) = 0 is -1 - m.
which we want = -1 / (½) or -1 - m = 2, or m = 1.
\The altitude thru P₁ is (x + y - 6) + 1 (x - 1) = 0,
i.e., L4: 2 x + y - 7 = 0.
     Similarly, the family of lines thru P₂ is
L_n: (2 y - x - 3) + n (x - 1) = 0,
which has slope (1 - n)/2.
L₂, which is y = - x + 6, has slope -1. \We want (1 - n)/2 = -1 / -1, or n = -1.
Then L_n becomes 2 y - x - 3 - 1 (x - 1) = 0,
i.e., L5: y = x + 1.
     The orthocenter H lies on L4 and L5. Solving them simultaneously by substituting
y = x + 1 into L4, we get
2 x + (x + 1) - 7 = 0, or x = 2. \ y = 3.
\The orthocenter H is (2, 3). QED

    Problem 6. Find the equations of a parabola K whose directrix is L1: 2x + y = 2 and whose focus is F(3, 2).
Solution. Let P(x, y) be any point on K, and N its projection on L1. Then PN = PF.
\| 2 x + y - 2 |/ Ö(2² + 1²) = Ö[(x - 3)² + (y - 2)²],
i.e., (2 x + y - 2)² / 5 = (x - 3)² + (y - 2)²,
i.e., 4 x² + y² + 4 + 4 x y - 8 x - 4 y = 5 [x² - 6 x + y² - 4 y + 13],
i.e., 0 = x² - 4 x y + 4 y² - 22 x - 16 y + 61 = 0,
which is the equation of the parabola. Notice (x² - 4 x y + 4 y²) is a perfect square.

    Problem 7. Find the equation of an ellipse K whose ratio of major and minor radii is 3 : 2 and whose foci are F(2,1) and F'(6,3).
    Solution. Let FF' = 2 c, then FF' = Ö[(6 - 2)² + (3 - 1)²] = Ö20 = 2 Ö5.
Let the major radius be a, the minor radius be b, then a/b = 3/2, and a² = b² + c².
Solving them, we get a = 3 and b = 2.
     Now if P(x, y) is any point on K, then PF + PF' = 2 a.
\Ö[(x - 2)² + (y - 1)²] + Ö[(x - 6)² + (y - 3)²] = 6,
i.e.,Ö[(x - 2)² + (y - 1)²] = - Ö[(x - 6)² + (y - 3)²] + 6.
\(x - 2)² + (y - 1)² = 36 + (x - 6)² + (y - 3)² - 12Ö [(x - 6)² + (y - 3)²].
\8 x + 4 y + 76 = - 12 Ö[(x - 6)² + (y - 3)²].
\ (2 x + y + 19)² = 9 [(x - 6)² + (y - 3)²],
\0 = 44 - 184 x - 92 y + 5 x² - 4 x y + 8 y² is the equation of the ellipse.

    Problem 8. A ladder has length a + b, and P is a point on it at distance b from its lower end on the floor. The top of the ladder leans on a vertical wall, and the bottom of the ladder is pulled away from the wall. Eventually the ladder falls flat on the floor. Find the trajectory (locus) of P.

    Solution. Let the wall be y-axis and the floor be x-axis.
Let the ladder be AB, A leaning on y-axis. Let M, N be the projections of P upon x-axis and y-axis. Then DPMB is similar to DANP.
Let P be (x, y), and q be the angle which the ladder makes with the wall.
Then sin q = x/a and cos q = y/b.
Eliminating q we get (x/a)² + (y/b)² = 1.
Hence P lies on the ellipse represented by this equation.

    Problem 9. Let A(5, 12), B(5, -12) be given points and
Ö8 y + x + 39 = 0 be a line L. Find the circles each of which going thru A and B and touching L.
    Solution. Let the center of one of the circles be P(p, q). Let its projection on L be N.
Then it lies on the perp. bisector of AB, which happens to be the x-axis.
\q = 0. We want PA = PB = PN.
\Ö[(p - 5)² + (0 - 12)²] = | Ö8×0 + p + 39 |/ Ö( Ö8² + 1²),
i.e., p² - 10 p + 13² = (p + 39)² / 9,
i.e., 8 p² - 168 p = 0. \ p = 0 or 21.
\The centers are (0, 0) or (21, 0).
The radii are Ö[(0 - 5)² + (0 - 12)²] or Ö[(21 - 5)² + (0 - 12)²],
i.e., 13 or 20.
Hence the equations of the circles are x² + y² = 13² and
(x - 21)² + y² = 20² .

    Theory of Quadratics. We need several lemmas.
    Lemma 1. Remainder Theorem. Let f(x) = a x² + b x + c be a quadratic polynomial over the real numbers.
Then for any r, dividing f(x) by x - r, the remainder is f(r).
Proof. Dividing f(x) by (x - r) will yield, by long division, a quotient p x + q and a remainder s.
Then
f(x) = (p x + q) (x - r) + s for all x.
Substitute r for x, we get f(r) = s. QED
    Lemma 2. Factor Theorem Let f(x) = a x² + b x + c be a quadratic polynomial over the real numbers. Then r is a root, iff (x - r) is a factor.
Proof. If r is a root, then f(r) = 0, hence from Lemma 1,
f(x) = (p x + q) (x - r) + s where s = f(r) = 0. Thus f(x) is factorizable as (p x + q) (x - r).
Conversely, if (x - r) is a factor, then f(x) = (p x + q) (x - r) for some p, q.
Hence f(r) = 0, and r is a root.
    Lemma 3. Unique Factorization of a quadratic polynomial with real coefficients. Let
a (x - b) (x - c) = a' (x - b') (x - c') . . . (e1), then a = a', and {b', c'} = {b, c}. Proof. If we expand both sides and compare the x² terms we see that a must = a'.
If b' ¹ b and b' ¹ c, then LHS of (c1) = a (b' - b) (b' - c) ¹ 0,
whereas RHS = 0, which is absurd, because (e1) is true for all x.
Hence it is impossible for b' to be different from b and c, and similarly for c'.QED
    Corollary. Hence any quadratic polynomial with real coefficients has at most 2 roots, possibly complex.
    Lemma 4. Quadratic Roots' sum and product.
Let f(x) = a x² + b x + c be a quadratic polynomial over the real numbers.
If m and n are roots, then m + n = -b/a and m n = c/a.
Proof. From Lemma 2, a x² + b x + c = a (x - m) (x - n). = a x² + a (m + n) x + a m n.
Comparing both sides, \ b = a (m + n), c = a m n. QED
    Lemma 5. If f(x) = a x² + 2 h x + c has a repeated root, then h² = a c, also the root is -h/a.
Proof. let b = 2 h. Then quadratic formula gives the roots as
[-b ±Ö(b² - 4 a c)] / 2 a = [-b ±Ö(h² - a c)] / a.
Repeated root means the Ö... part must be zero, hence h² = a c, and the root is - h / a. QED

    Problem 10. Let P(-a, b) be a point on the directrix x + a = 0 of the parabola
K: y² = 4 a x . . . . (e1).
Prove that the two tangents from P touch K at two points A, B such that PA ^ PB and that AB passes through the focus F(a, 0).
    Solution. A line thru P is L: (y - b) = m (x + a) . . . . (e2)
where m is the slope. It intersects K at points (x, y) satisfying both equations.
Eliminating y, we get (b + m x + m a)² = 4 a x, or
m² x² + (2 m (m a + b) - 4 a) x + (m a + b)² = 0 . . . . (e3)
For L to be a tangent, (e3) must have a repeated root for x.
\ by Lemma 5, (m (m a + b) - 2 a)² = m² (m a + b)²,
i.e., - 4 m a (m a + b) + 4 a² = 0, i.e., m (m a + b) = a,
i.e., a m² + b m - a = 0 . . . . (e4).
As a quadratic equation in m, the product of its roots m₁ and m₂ is - a / a = -1.
Hence the two tangents are PA: y - b = m₁ (x + a) and
PB: y - b = m₂ (x + a) where the product of their slopes is -1, proving PA ^ PB.
Let A be (x, y) which satisfies (e1) and (e3), and m satisfies (e4). Using (e4), (e3) simplifies to
m² x - 2 a x + a² / m² = 0,
i.e., (m x - a/m)² = 0,giving x = a / m².
From (e1), y = Ö(4 a x) = 2 a / m.
\ y/(x - a) = 2 m / (1 - m²) = from (e4), 2 a / b, proving that both A and B lies on the line
y = (2 a / b) (x - a), which goes thru the focus (a, 0). QED

    Problem 11. Find the circles each going thru the point P(0, 8), touching the x-axis, and touching the circle
C2: x² - 30 x + y² - 28 y + 321 = 0 . . . . (e2)
    Solution. C2 is (x - 15)² + (y - 14)² = 10²,
which has (15, 14) as center O₂ and radius 10.
Let Q(x, y) be the center of the sought circles, with radius r.
Then The distance of Q from x-axis is r, i.e., y = r . . . . (e3).
Moreover, PQ = r, so that (x - 0)² + (y - 8)² = r²,
i.e., x² = 18 r - 81, or r = x²/18 + 9/2 . . . . (e4).
Finally, QO₂ = r + 10, so that (x - 15)² + (y - 14)² = (r + 10)²,
which by (e3) becomes x² - 30 x + 321 = 48 r,
which by (e4) becomes x² - 30 x + 321 = 8 x²/3 + 216.
\ (5/3)x² + 30 x - 105 = 0, or x² + 18 x - 63 = 0.
\ (x + 21) (x - 3) = 0. \ x = -21 or 3, and then y = r = 29 or 5.
\One circle is (x - 3)² + (y - 5)² = 25, and the other is
(x + 21)² + (y - 29)² = 29².

    Exercises.
    Ex. 1. Find a circle centered at A(2, 0)and touching the line (e1): 2 x + y = 3.
    Ex. 2. Given a curve y² = 4 a x. Prove it is a parabola by proving any point P(x, y) on it is equidistant from the focus F(a, 0) and the directrix x + a = 0.
    Ex. 3. Prove that the tangents at the ends of a focal chord of a parabola intersect at right angles on the directrix.
    Ex. 4. Let P(x', y') be a fixed point on the parabola y² = 4 a x.
Prove that the tangent there makes equal angles with PF and PN, where F is the focus (a, 0) and N is the projection of P on the directrix.
    Ex. 5. Define the ellipse as the locus of points whose sum of distances from
F(c, 0) and F'(-c, 0) = 2 a, for some constant a > c.
Prove it has equation x²/a² + y²/b² = 1, where b² + c² = a².
Prove without coord geometry if possible, that if e = c/a, and L is the line x = a/e,
then for all points P on the ellipse, PF = e PN, where N is the projection of P on L.
    Ex. 6. Let L be a fixed line and F a fixed point not on L, and e some fixed number between 0 and 1. Let K be the locus of points P satisfying FP = e NP, where N is the projection of P on L. Prove K is an ellipse as defined in Ex. 5.
    Ex. 7.Prove that 2 x² - 5 x y + 2 y² - 2 x + 7 y - 4 = 0 is two str. lines by factorizing it. Find their intersection point.
    Ex. 8.Find the equation of a parabola with axis being x + 1 = y, and going through the points (0, 2) and (0, 4).
    Ex. 9.Find the area of the triangle with vertices (1, 1), (2, -1), (-1, 3). Find its circumcircle without first finding its center. Find its incenter and incircle.

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