Find the distance from (-3,1) to line AB: A(-6,14) B(2,4)?

Question

help quick

Answer 1

the distance from a point to a line means the perpendicular distance is to be calculated.
the slope of the line AB , m = (y2 - y1)/(x2 - x1)
= (4 - 14)/(2 -(-6)
= -10/8 = - 5/4
the equation line AB is y - 4 = -5/4(x - 2)
4y - 16 = -5x + 10
4y + 5x = 26 ------------ eqn (1)
the slope of the line perpendicular to AB = 4/5(since m1*m2 =1)
let the equation of the line CD perpendicular to AB =
(y -k) = m(x - h)
where k = 1, h = -3 and m = 4/5
y - 1 = 4/5(x + 3)
5y - 5 = 4x + 12
5y - 4x = 17 ---------- eqn (2)
let the coordinates of the point D where it meets AB are x3, y3. This point D is common to both AB and CD
so substitute x3, y3 in eqn 1) and 2)
4y3 + 5x3 = 26 -------- eqn (3)
5 y3 - 4 x3 = 17 ------- eqn (4)
multiply eqn(3) by 5
20 y3 + 25 x3 = 130 ----- eqn (5)
multiply eqn(4) by 4
20 y3 - 16x3 = 68 ------ eqn(6)
subtract eqn(6) from eqn(5)
41 x3 = 62
x3 = 62/41
substitute this in eqn(3)
4y3 + 5(62)/41 = 26
4y3 = 26 - 310/41
=756/41
y3 = 756/(4)(41)
= 189/41
the distance between(-3, 1) and (62/41, 189/41)
sqrt[(62/41 +3)^2 + (189/41-1)^2]
sqrt[(185/41)^2 + (148/41)^2]
sqrt[34225 + 21904/(41)^2]
sqrt[56129]/41 = 237/41=5.8 units

Answer 2

Find the distance from (-3,1) to line AB: A(-6,14) B(2,4).

First find the equation of line AB. Calculate the slope m.

m = ∆y/∆x = (4 - 14)/(2 + 6) = -10/8 = -5/4

Now plug in a point to write the equation of the line. Let's choose point B(2, 4).

y - 4 = (-5/4)(x - 2)

Multiply by 4 to eliminate fractions.

4y - 16 = -5(x - 2)
4y - 16 = -5x + 10
5x + 4y - 26 = 0

Now use the distance formula.

d = | 5*(-3) + 4*1 - 26 | / √(5² + 4²)
d = | -15 + 4 - 26 | / √(25 + 16)
d = | -37 | / √41

d = 37 / √41 ≈ 5.7784292
__________

Answer 3

many ways lead to Rome. Here is one of them:

apply vector analysis:
1. write the given points position vectors with their components in x and y.
r=(rx,ry)
rA=(rAx,rAy)=(-6,14)
rB=(rBx,rBy)=(2,4)
rQ=(rQx,rQy)=(-3,1)

2. calculate the difference vector a:
(a is the vector which starts in the point A end ends in B)

a=rB-rA=(rBx-rAx , rBy-rAy)=(2-(-6) , 4-14)=(8,-10)

3. calculate the amount of the vector a: (with Pythagoras)
I a I= sqrt ( (rax)^2 +(ray)^2 )=sqrt ( (8)^2+(-10)^2 )=sqrt(164)

4. calculate the difference vector (rQ-rA):
( (-3)-(-6), 1-(-14) ) = (3,-13)

5. calculate the VECTOR-Product of a x (rq-rA) :
Note: the vector-product for the vectors a and (rQ-rA) will lead to a determinant
Determinant D=8*(-13) - (-10*3) = -74

6. calculate the amount of the determinat D:
I D I = I a x (rQ-rA) I = 74

7 calculate the distance d from the point Q to the line AB:

d= I D I / I a I = 74/sqrt (164) = 5.78
d=5.78