maths help?

Distance between the point (x, -1) and the line 3x - 4y + 7 = 0 :

|3x - 4(-1) + 7| / √[3² + (-4)²] = 8

|3x + 11| / 5 = 8

|3x + 11| = 40

3x + 11 = 40 or 3x + 11 = -40

3x = 29 or 3x = -51

x = 29/3 or x = -17

The perpendicular distance from (x,-1) to 3x-4y+7 is
|3x+4+7|/sqrt(3^2+4^2)=|3x+11|/5=8 so
|3x+11|=40 so
3x+11=40 giving x=29/3 OR
3x+11=-40 giving x=-55/3

(xA ; - 1) ← this is the point A

y = - 1 ← this is the line (ℓ1) ← in blue


3x - 4y + 7 = 0 ← this is the line (ℓ2) ← in red

4y = 3x + 7

y = (3/4).x + (7/4) ← the slope of the line (ℓ2) is (3/4)



You have to find the line (ℓ3) that is perpendicular to the line (ℓ2).

The typical equation of a line is: y = mx + b → where m: slope and where b: y-intercept

Two lines are perpendicular if the product of their slope is: - 1

The slope of the line (ℓ2) is (3/4), so the slope of the line (ℓ3) is - (4/3).

The equation of the line (ℓ3) becomes: y = - (4/3).x + b

You know that the line (ℓ3) passes through the point A, so the coordinates of this point A must verify the equation of the line (ℓ3).

y = - (4/3).x + b

b = y + (4/3).x → you substitute x and y by the coordinates of A (xA ; - 1)

b = - 1 + (4/3).xA


The equation of the line (ℓ3) is:

y = - (4/3).x + [- 1 + (4/3).xA]

y = - (4/3).x + (4/3).xA - 1 ← this is the line (ℓ3)



The intersection between the line (ℓ3) and the line (ℓ1) gives the point A. → A (xA ; - 1)

The intersection between the line (ℓ3) and the line (ℓ2) gives the point B.

…and you know that: AB = 8



Intersection between (ℓ3) and (ℓ2) gives B.

(ℓ3) : y = - (4/3).x + (4/3).xA - 1

(ℓ2) : y = (3/4).x + (7/4)


y = y

- (4/3).x + (4/3).xA - 1 = (3/4).x + (7/4)

- (4/3).x - (3/4).x = (7/4) - (4/3).xA + 1

- (16/12).x - (9/16).x = (21/12) - (16/12).xA + (12/12)

- 16x - 9x = 21 - 16.xA + 12

- 25x = 33 - 16.xA

x = (16.xA - 33)/25 ← this is xB → i.e. the abscissa of the point B

xB = (16.xA - 33)/25


Recall (ℓ2)

y = (3/4).x + (7/4) → where: x = xB

y = (3/4).xB + (7/4) → xB = (16.xA - 33)/25

y = (3/4).[(16.xA - 33)/25] + (7/4)

y = [(48.xA - 99)/100] + (175/100)

y = (48.xA - 99 + 175)/100 ← this is yB → i.e. the ordinate of the point B

yB = (48.xA + 76)/100



Distatnce AB

xAB = xB - xA = [(16.xA - 33)/25] - xA = (16.xA - 33 - 25.xA)/25 = (- 9.xA - 33)/25

yAB = yB - yA = [(48.xA + 76)/100] + 1 = (48.xA + 76 + 100)/100 = (48.xA + 176)/100

AB² = xAB² + yAB²

AB² = [(- 9.xA - 33)/25]² + [(48.xA + 176)/100]² ← you know that: AB = 8 → AB² = 64

[(- 9.xA - 33)/25]² + [(48.xA + 176)/100]² = 64

[4.(- 9.xA - 33)/100]² + [(48.xA + 176)/100]² = 64

[16.(- 9.xA - 33)²/100²] + [(48.xA + 176)²/100²] = 64

16.(- 9.xA - 33)² + (48.xA + 176)² = 100² * 64

16.(81.xA² + 594.xA + 1089) + (2304.xA² + 16896.xA + 30976) = 100² * 64

1296.xA² + 9504.xA + 17424 + 2304.xA² + 16896.xA + 30976 = 100² * 64

3600.xA² + 26400.xA + 48400 = 100² * 64

36.xA² + 264.xA + 484 = 100 * 64

9.xA² + 66.xA + 121 = 100 * 16

9.xA² + 66.xA - 1479 = 0

3.xA² + 22.xA - 493 = 0


Polynomial like: ax² + bx + c, where:
a = 3
b = 22
c = - 493

Δ = b² - 4ac (discriminant)

Δ = 22² - 4.(3 * - 493)

Δ = 6400 = 80²



xA = (- b - √Δ)/6 = (- 22 - 80)/6 = - 17 ← this is the first possibility for the abscissa of A

xA = (- b + √Δ)/6 = (- 22 + 80)/6 = 29/3 ← this is the second possibility for the abscissa of A


→ A1 (- 17 ; - 1) ← this is the first point A

→ A2 (29/3 ; - 1) ← this is the second point A



You can graph the situation.

(xA ; - 1) ← this is the point A

y = - 1 ← this is the line (ℓ1) ← in blue


3x - 4y + 7 = 0 ← this is the line (ℓ2) ← in red



You draw the point A1 (- 17 ; - 1).

You draw the circle with its center @ A1 and a radius r1 = 8.

You can see that the circle is tangent to the line (ℓ2) @ B1.



You draw the point A2 (29/3 ; - 1).

You draw the circle with its center @ A2 and a radius r2 = 8.

You can see that the circle is tangent to the line (ℓ2) @ B2.