Chapter 2 Analytic Geometry Distance and Lines

50 AMC Lectures Chapter 17 Analytic Geometry Distance and Lines

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BASIC KNOWLEDGE 1. Distance formula The distance (d) between two points and can be calculated by the following formula:

(1) Proof: Given right triangle ABC, from the Pythagorean Theorem, we have

Let AB = d,

Taking the square root on both sides:

Another useful form of the distance formula:

= = (2)

Where k is the slope of the line containing the two points and .

2. Finding the coordinates of a point P on between points A and B.

P is a point on between points A and B. If , then we have the following

formulas to calculate the coordinates of P:

, and (3)


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When λ = 1, we get the midpoint formula: The coordinates of the midpoint of the line segment with endpoints and

are , and . (4)

3. Point to line distance formula (1). The distance from a point (x1, y1) to a line ax + by +c = 0 is one of those problems that seems easy, but is very time consuming, unless you know the following formula:

(5)

Proof: Method 1: Drop perpendiculars to the line l from the given point to

meet l at , to the x-axes to meet l at ,

and to P0Q from N to meet P0Q at .

The triangle P0QN in the picture is a right triangle, so the area can be found in two ways:

, or

.

Solving for d:


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Method 2: Find the distance from the point to the line . meets x-axes and y-axes at

, respectively.

Drop perpendiculars to the line l from the given point

to meeting l at , to the x-axes to meet

l at ,

So Rt∆P0QN ~ Rt∆EFO.

⇒ ⇒

(2). Other forms of the formula of point to distance

(i). (6)

It tells us that the distance from a point (x0, y0) to the line l: ax + by = 0 is less than or equal to the distance from the point to origin if the line l passes through the origin and the point is not on the line.


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Note that if we square both sides of (6), we get a famous inequality: Cauchy Inequality: .

(ii). (7)

It tells us that all the line segments connecting a point that is not on the line l: ax + by + c = 0 to a point that is on the line l, the perpendicular is the shortest. (3). If line l1 and line l2 are parallel, the distance between

them is d and (8)

4. Angle formed by two lines The slopes of line l1 and line l2 are k1 and k2, respectively. The angle formed by lines l1 and l2 is θ and

tan θ = (9)

When , θ = .

The slopes of line l1 and line l2 are k1 and k2, respectively. The angle from lines l1 to l2 is θ and

(10)

When , θ = .

5. Pattern in Reflections (1). is a point. The image of P under reflections:

(a). In the x−axes

(b). In the y−axes


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(c). In the x = a

(d). In the y = a

(e). In the y = x

(f). In the y =− x

(g). In the y = x + m

(h). In the y = − x + n

(i). In the point A(a, b)

(j). In the Ax + By +C = 0

The figures for (a) to (f) are showing below.

(a) (b) (c)

(d) (e) (f) (2). The reflection of the line Ax + By +C = 0 in the point P( a, b ): Ax + By – (2aA + 2bB +C) = 0


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Example 1: Find the image of P(– 5,13) under the reflection in line 2x – 3y – 3 = 0. Solution: Let the image be Q and R be the midpoint of PQ.

So .

The slope of the line PQ is .

By the formula (2), we have

Therefore ⇒ .

We also have: ⇒ .

So the image Q is (11, –11). Example 2: A line passing through (1, 2) is cut by two parallel lines 4x + 3y + 1 = 0 and 4x + 3y + 6 = 0. The length of the segment cut is . Find the equation of the line. Solution: Let k be the slope of the line. Let the line be , and the two intersecting points with two parallel lines be

and .

⇒ .


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⇒ .

Since .

Simplifying we get:

Solve for k: , .

The line seeking is and .

Or x + 7y – 15 = 0 and 7x – y – 5 = 0. Example 3: Show that and if with . x, y, a, b, and c are real numbers. Solution: Rewrite the given equation as

This is the distance from (0, 0) to the line . Points (a, b) and (x, y) are on the line. Therefore or , . Example 4: Find the smallest value of if x – 2y + 2 = 0. Solution: From , we have . Examining the point (1, – 2) and the line x – 2y + 2 = 0:

(1)


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Squaring both sides of (1): .

That is ⇒ .

The smallest value of is achieved when y = −2 and x = .

Example 5: Line always contains the point P for any real value of m. What is the coordinates of the point P? Solution: (9, – 4). Write the given equation in the following form: For any real number of m, the line always goes through the point of intersection of lines

and . Solving x and y:

⇒ x = 9 and y = – 4.

The answer is: (9, – 4). Example 6: If the y-intercept of the line is 1, what is the value of the real number m? Solution: 3. Since the y-intercept is 1, (0, 1) is on the line. Substitute (0, 1) into the equation, we have:

⇒ Solving for m: m = 3 or m = – 1. If m = – 1, we have m + 1 = 0 and is not a line. So m = 3 is the answer. Example 7: Find the distance between the point (2, – 3) and the line 3x − 4y −12 = 0. Solution: 6/5.


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By the point to line distance formula:

Example 8: What is the closest that the line 4x + 5y = 20 comes to the origin? Solution:

Since 4x + 5y − 20 = 0, we have

Example 9: Find the distance between the parallel lines 2x − 3y = 12 and 2x – 3 y = 36. Solution: Pick any point on one line and plug it into the distance formula for a point and a line. So let’s pick (6, 0) from the first line. Now

Example 10: (2003 Duke Math Meet) What is the closest that the line comes

to a lattice point? Solution: The line, in standard form, is 20x − 35y + 7 = 0 , so the distance to any lattice point (x, y)

is .

So the shortest distance will be obtained when the numerator is minimized. Since the distance is an absolute value, and the coordinates of the point must be integers, we need to see if this quantity can be zero, or if not 0, what is the smallest possible value. Clearly 20x − 35y will always be a multiple of 5, so we could make it – 5, making the entire numerator 2. This is the smallest possible value for the numerator, so the shortest distance

is


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Example 11: P(x, y) is a point on a circle of radius of 1. If the center of the circle is (– 2,

0), what is the smallest value of ?

Solution: .

Let . It is clear that k is the slope of the line connecting points P(x, y) and A(2,

1). The smallest value of is the smaller one of the two slopes of the tangent lines to

the circle. The equation of the line AP is By the point to line distance formula, we have:

⇒

Example 12: What is the shortest distance between the circle x2 + y2 = 25 and the line 3x + 4y = 48?

Solution: .

First notice that the closest the line gets to the origin is , so it is

more than 5 units from the origin. Now subtract the radius of the circle, yielding the

distance

Example 13: Find tangent of the angle formed by two lines both passing through the origin. It is known that these two lines trisect the segment in the first quadrant of the line 2x + 3y = 12. Solution: Let A(6, 0) and B(0, 4).


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Since , .

Since ,.

∴C (2, ), D(4, ). , ⇒ .

Example 14: Find the smallest possible value of . Solution:

That is, to find the smallest possible sum of the distances from (x, 0) to points (1, 1), (2, – 2). That is, the distance between (1, 1) and (2, – 2), which is . Example 15: As shown in the figure, triangle ABC has the vertices A(x1, y1), B(x2, y2), and C(x3, y3). Find the coordinates of G, the centroid (the point where three medians meet). Solution: Let the coordinates of G be G(x, y). Let D be the midpoint of BC. By the midpoint formula, we get the coordinates of D:

.

Since G is the centroid of the triangle ABC, we know that AG : GD = 2 : 1.

That is .


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By the formula , and , we find the coordinates of G:

, and

.

Example 16: The vertices are A(4, 1), B(7, 5), and C(−4, 7) of triangle ABC. Find the equation of the angle bisector of ∠A. Solution: Let p(x, y) be a point on the angle bisector AD. Method 1: The distance from p to AC is the same as the distance from p to AB.

The equation for AB: or .

The equation for AC: or .

or .

It is easy to know that is the equation of the angle bisector of the exterior angle of ∠A. The equation is then . Method 2:

By the angle bisector theorem,


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So , then:

, and

The equation is:

⇒ ⇒

⇒ .


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PROBLEMS

Problem 1: What is the closest that the line comes to a lattice point?

Problem 2: (1966 AMC) Let m be a positive integer and let the lines 13x + 11y = 700 and y = mx – 1 intersect in a point whose coordinates are integers. Then m can be: (A) 4 only (B) 5 only (C)6 only (D) 7 only (E) one of the integers 4, 5, 6, 7 and one other positive integer Problem 3: (1982 AMC) A vertical line divides the triangle with vertices (0, 0), (1,1) and (9,1) in the xy-plane into two regions of equal area. The equation of the line is x = (A) 2.5 (B) 3.0 (C) 3.5 (D) 4.0 (E) 4.5 Problem 4: (1997 NC Math Contest) Let P be the point (3,2). Let Q be the reflection of P about the x-axis, let R be the reflcetion of Q about the line y = – x and let S be the reflection of R through the origin. Then PQRS is a convex quadrilateral. What is the area of PQRS? (a) 14 (b) 15 (c) 16 (d) 17 (e) 18 Problem 5: (2001 NC Math Contest) Find the slope of the line with a positive rational slope, which passes throug the point (6, 0) and at a distance of 5 from (1, 3). Write the

slope in the form , where a and b are relatively prime. What is the sum of a and b?

a. 24 b. 23 c. 22 d. 21 e. none of these. Problem 6: Line 1 (l1) x – 2y + 3 = 0 intersects x-axes at A. Line 2 (l2) is obtained by rotating l1 45° about A. If the rotation is counterclockwise, find the equation for l2. Problem 7: Find a + b if the line ax + y + 1 = 0 is parallel to the line (a + 1)x + by + 2 = 0 and two lines have the same distance to the origin.

(A) (B) 3. (C) or 3. (D) any value except 3,

Problem 8: Find the equation of the line passing through p(1, 2) and the length of its segment cut between two lines and is .


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Problem 9: Find the image of P(−2, −1) under the reflection in line . Problem 10: Find the image of line under the reflection in line

. Problem 11: Find the image of the line under the reflection in (1, 1). Problem 12: Find the greatest possible value of .

Problem 13: Find the equation of the line passing through point P(−5, 4). It is known that P divides the segment A(x, 0) and B(0, y) of the line between the x and y axis into the ratio of 1:2. Problem 14: A line segment is between two lines: and

. The midpoint of the segment is P(0, 1). Find the equation of the line containing the segment. Problem 15: Find the equation of the angle bisector of the angle formed by lines

and . Problem 16: Find a if the distance between line l1: 2x + 3y – 6 = 0 and line l2: 4x + 6y +

a = 0 is .

Problem 17: Find the equation of the line passing through point P(2, −1) and having a distance 2 from the origin. Problem 18: (2001 AMC 12) Points A = (3, 9), B = (1, 1), C = (5, 3), and D = (a, b) lie in the first quadrant and are the vertices of quadrilateral ABCD. The quadrilateral formed by joining the midpoints of , , , and is a square. What is the sum of the coordinates of point D? (A) 7 (B) 9 (C) 10 (D) 12 (E) 16


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Problem 19: (2003 AMC 12 A) A set S of points in the xy-plane is symmetric about the origin, both coordinate axes, and the line y = x. If (2, 3) is in S, what is the smallest possible number of points in S? (A) 1 (B) 2 (C) 4 (D) 8 (E) 16 Problem 20: (2004 AMC 12 A) Let A = (0, 9) and B = (0, 12). Points and are on the line y = x, and and intersect at C = (2, 8). What is the length of

€

A'B'? (A) 2 (B) (C) 3 (D) (E) Problem 21: (2004 AMC 12 B) The point (- 3, 2) is rotated 90o clockwise around the origin to point B. Point B is then reflected in the line y = x to point C. What are the coordinates of C? (A) (3, 2) (B) (2, 3) (C) (2, 3) (D) (2, 3) (E) (3, 2) Problem 22: (2000 AMC12) The point P = (1, 2, 3) is reflected in the xy-plane, then its image Q is rotated by 180° about the x-axis to produce R, and finally, R is translated by 5 units in the positive-y direction to produce S. What are the coordinates of S? (A) (1, 7, – 3) (B) (–1, 7, –3) (C) (–1, –2, 8) (D) (–1, 3, 3) (E) (1, 3, 3) Problem 23: (1981 AMC 12) The lines L and K are symmetric to each other with respect to the line y = x. If the equation of line L is y = ax + b with a ≠ 0 and b ≠ 0, then the equation of K is y =

(A) (B) (C) (D) (E)

Problem 24: (2001 NC Math Contest Algebra II) Given A(1,− 2) , B(5, 1) , and C(−2, 2) , find the equation of the angle bisector at A. a. 5x − y = 7 b. 7x − y = 2 c. 7y − x = 2 d. y = x + 3 e. none of the above


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Problem 25: (1980 AMC 12) The equations of L1 and L2 are y = mx and y = nx, respectively. Suppose L1 makes twice as large an angle with the horizontal (measured counterclockwise from the positive x-axis) as does L2, and that L1 has 4 times the slope of L2. If L1 is not horizontal, then mn is

(A) (B) (C) 2 (D) – 2

(E) not uniquely determined by the given information Problem 26: (1990 AIME) A triangle has vertices P = (– 8, 5), Q = (–15, –19) and R = (1, –7). The equation of the bisector of can be written in the form ax + 2y + c = 0. Find a + c.


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SOLUTIONS TO PROBLEMS Problem 1: Solution: Let m and n are the coordinates of a lattice point. The distance from this lattice point (m, n) to the given line is

We know that is a multiple of 7. We also know that 3 and 5 are relatively prime. Since we want to get the smallest value for d, or the smallest value for the numerator, so = −7 is the best value we could get (m = – 2 and n = – 1).

Problem 2: Solution: (C). Substituting y from the second equation into the first gives 13x + 11(mx – 1) = 700, so that

Since x is to be an integer, the denominator 13 + 11m must be a divisor of the numerator, and its only divisors are 1, 3, 32, 79, 3·79, 32·79. Our task now is to find a positive integer m such that

13 + 11m = d, or

Where d is one of these divisors. Since m > 0, we see that d >13, so the only divisor we need to test are the last three:

(i) if d = 79, d – 13 = 66, and

(ii) if d = 3·79 = 237, d – 13 = 224 is not divisible by 11 (iii) if d = 32·79 = 711, d – 13 = 698 is not divisible by 11. We conclude that m = 6 is the only positive integer yielding a lattice point for the intersection of the given lines.


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Problem 3: Solution:(B). Method 1 (official solution): In the adjoining figure, ABC is the given triangle and x = a is the dividing line.

Since area the two regions must

each have area 2. Since the portion of

∆ABC to the left of the vertical line through vertex A has area less than area

the line x = a is indeed right of A as shown. Since the equation of line BC is the

vertical line x = a intersects BC at a point E: Thus

area or (9 – a )2 = 36.

Then 9 – a = ± 6, and a = 15 or 3. Since the line x = a must intersect ∆ABC, x = 3. Method 2 (our solution): Extend CA to meet y-axes at D. We see that ∆CDA ∼ ∆CGE. Therefore, we get:

⇒ ⇒

We also see that the area of ∆CDA = the area of ∆CAB + the area of at ∆DAB = the area of ∆CGE × 2 + the area of at ∆DAB.

⇒

⇒ ⇒ ⇒

⇒ . DG = DC – CG = 9 – 6 = 3. Therefore x = 3. Problem 4: Solution: The points P, Q, R, and S are respectively (3, 2), (3, – 2), (2, – 3) and ( – 2, 3). The


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rectangle with vertices( – 2, 3), (3, 3), (3, – 3) and ( – 2, – 3 ) has the area of 30. When

the three corner regions, with the areas of , and 12, respectively, are subtracted, the

remaining quadrilateral has the area of 15. Problem 5: Solution: b.

, , .

.

. .

Answer: 8 + 15 = 23. Problem 6: Solution: Let the slope of l2 be k and l1 be α.

We have and .

Since the intersecting point is A(– 3, 0) , the equation of line 2 is then y = 3(x + 3) or 3x – y + 9 = 0. Problem 7: Solution: (A). Since two lines are parallel, ab = a + 1 (1) Since they have the same distance to the origin,

(2)

Solving the system of equations (1) and (2) gives:

or

When a = 1, b = 2, these two lines are overlapped.


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Therefore b = – 2 ⇒

Problem 8: Solution: Method 1: Let the equation be . The points of intersection with two lines are A and B, respectively. Solving the system equations of

and

The coordinates of A and B are:

and .

Since , .

The above equation has the form of: (7k + 1)(k – 7) = 0 ⇒ or 7.

So the equation is or . Method 2:

The distance between two parallel lines is .

Since the length of the segment cut by them is , the angle formed the line in question and two parallel lines is 45°. Let the equation be .


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⇒

Solving for k: .

So the equation is or . Problem 9: Solution: Let the image be Pʹ′ (x, y).

⇒

Solve for x and y: ⇒ .

Problem 10: Solution: Method 1:

Solving gives the point of intersection:

Let the slope of the line in question be k:

Solve for k: k = – 7 or k = 1(extraneous).

So the equation of the line is or .

Method 2: Since is a point on , its image is .


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Solve for x: or (extraneous).

So

The equation is .

Problem 11: Solution: Let the equation we want to find be We know that d, the distances from (1, 1) to these two parallel lines are the same.

.

Therefore

Solving for b: b = 0 or b = – 4 (extraneous). So the equation is y = 3x. Problem 12: Solution:

To find the greatest difference of the distances from P(x, 0) to A (1, 2) and P to

B(2, 3). The desired point P must be the point of intersection of the line segment connecting A and B and x-axes.


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Problem 13: Solution:

Let the equation of the line be: . The coordinates of A and B are A(a, 0), B(0,

b.).

Since , considering P(−5, 4), by formula (3), we obtain: , b = 12.

The equation is . Problem 14: Solution: Let the equation be y = kx +1. The points of intersection of line y = kx + 1 with

and are and ,

respectively.

By the midpoint formula (4), we get .

The equation is or .

Problem 15: Solution: Let P(x, y) be a point on the angle bisector. The distances from point P to two given lines

are the same. So .

The equations are: and . Problem 16: Solution:

We know that the distance between l1: 2x + 3y – 6 = 0 and l2: l2: 2x + 3y + = 0 is .

By the formula (8) , we have: .

Therefore . Solving for a gives a = – 7 or – 17.

Problem 17: Solution: If the slope of the line does not exist, the equation of the line is x = 2.


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If the slope of the equation exists, let it be k. The equation is then written as y + 1 = k(x – 2), or kx – y – 2k – 1 = 0.

By the point to line distance formula, we have ＝2.

Solve for k: .

The equation of the line is: or 3x – 4y – 10 = 0.

The equations are x – 2 = 0 or 3x – 4y – 10 = 0. Problem 18: Solution: (C). Let the midpoints of sides , , , and be M, N, P, and Q, respectively. Then

M = (2, 5) and N = (3, 2). Since has slope – 3, the slope of must be , and MQ

= MN = . An equation for the line containing is thus or

Therefore Q has coordinates of the form (a, Since , we

have ⇒ ⇒

⇒ ⇒ .

We know that Q is in the first quadrant, so a = 5 and Q = (5, 6). Since Q is the midpoint of and A = (3, 9), we have D = (7, 3), and the sum of the coordinates of D is 10. Problem 19: Solution: (D). The set S is symmetric about the line y = x and contains (2, 3), so it must also contain (3, 2). Also S is symmetric about the x-axis, so it must contain (2, – 3) and (3, – 2). Finally, since S is symmetric about the y-axis, it must contain (– 2, 3), (– 3, 2), (– 2, – 3), and (– 3, – 2). Since the resulting set of 8 points is symmetric about both coordinate axes, it is also symmetric about the origin.


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Problem 20: Solution: (B).

Line AC has slope and y-intercept (0, 9), so its equation is

Since the coordinates of Aʹ′ satisfy both this equation and y = x, it follows that Aʹ′ = (6, 6). Similarly, line BC has equation y = 1 – 2x + 12, and Bʹ′ = (4, 4).

Thus ʹ′Bʹ′ . Problem 21: Solution: (E). The rotation takes (- 3, 2) into B = (2, 3), and the reflection takes B into C = (3, 2). Problem 22: Solution: (E). Reflecting the point (1, 2, 3) in the xy-plane produces (1, 2, –3). A half-turn about the x-axis yields (1, –2, 3). Finally, the translation gives (1, 3, 3). Problem 23: Solution: (E). If (p, q) is point on line L, then by symmetry (q, p) must be a point on K. Therefore, the points on K satisfy x = ay + b.

Solving for y yields .

Problem 24: Solution: (e). We find that side AC and AB both have length 5, so the angle bisector is also the median, passing through A(1, −2) and the midpoint (3/2, 3/2), of side BC. This makes the slope 7 and the equation y = 7(x − 1) − 2 = 7x − 9.


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Problem 25: Solution: (C). In the adjoining figure, L1 and L2 intersect the line x = 1 at B and A, respectively; C is the intersection of the line x = 1 with the x-axis. Since OC = 1, AC is the slope of L2 and BC is the slop of L1. Therefore AC = n.

Since OA is an angle bisector, .

This yields and OB = 3.

By the Pythagorean theorem so .

Problem 26: Solution: Method 1: Let the bisector be PT and the coordinates of T be (x, y).

,

By formula (3), we have x = – 5, .

The equation for PT: or 11x + 2y + 78 = 0.

Therefore a + c = 11 + 78 = 89. Method 2: Let the slope of the angle bisector be k. The slopes of PR and PQ are kPR and kPQ, respectively.

By the formula (1), we have:

⇒ 22k2 + 117k – 22 = 0 ⇒ .


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Since the angle bisector is the interior angle bisector, .

or 11x + 2y + 78 = 0.

Therefore a + c = 11 + 78 = 89.

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Chapter 2 Analytic Geometry Distance and Lines