what is the distance from the point (2,6) to 2x-y=3 ??? PLEASE HELP. dont know if i did it right!?

Question

is this right?

2x-y=3
-2x -2x
-y= -2x+3
-1(-y=-2y+3)
y= 2x - 3

the perpendicular line that would interesect that line at that point would be:
y = -1/2x +6

intersection (4.25, 5.5)
distance = √ 5.3125

is this right?

Answer 1

The equation of the line is:
y = 2x - 3

So the slope is 2. The perpendicular slope would be the negative reciprocal -½.

The general equation of this new line will be.
y = -½x + k

Now plug in the point (2, 6) to figure out the perpendicular line.
6 = -½(2) + k
6 = -1 + k
k = 7

This is apparently where you made an arithmetic error. The correct formula for the perpendicular line is:
y = -½x + 7

The point where these lines intersect is found by equating the two lines:
y = 2x - 3
y = -½x + 7
2x - 3 = -½x + 7
2½x = 10
x = 10 / 2.5
x = 4
y = 2(4) - 3
y = 5

So the intersection point is (4, 5)

The distance between (2,6) and (4,5) is:

D = sqrt((4-2)² + (5-6)²)
D = sqrt(4 + 1)
D = sqrt(5)

So the distance is sqrt(5) ≈ 2.23606798

I think you had the right method but just made one slight mistake at step 2.

Answer 2

The distance from a point to any line is the perpendicular distance from the point to the point of intersection of the line and perpendicular drawn from the point to the line.

let the point of intersection of the perpenducular from point(2,6) to the line 2x - y = 3 is (x1,y1)

As point (x1,y1) lies on the line

2x1 - y1 = 3 -----------------eqn(1)

find out the the slope of the given line by arranging the line in standard form

2x - y = 3 => y = 2x - 3

so slope is 2

so the slope of the perpendicular line from the point(2.6)

is = -1/2

But the slope of the line joining (2,6) and (x1,y1)

is

(y1- 6)/(x1- 2)

so (y1- 6)/(x1- 2) = -1/2

2y1 - 12 = -x1 + 2

x1 + 2y1 = 14 -------------eqn(2)

multiplying eqn(2) with 2

2x1 + 4y1 = 28 ---------eqn(3)

subtracting eqn(3) from eqn(1) 2x1 - y1 = 3

5y1 = 25

y1 = 5

substituting y1 in eqn (1)

2x1 - 5 = 3

2x1 = 8

x1 = 4

so the distance between (2,6) and (4,5)

sqrt[(2-4)^2 + (6-5)^2]

=>sqrt(4 + 1) = sqrt(5)

Note: you can find out the perpendicular distance from the point (p,q) to the line ax+by+c by the formula

d = [ap +bq +c]/sqrt(a^2+b^2)

here a = 2, b= -1 , c = -3 , p = 2 and q = 6

d = [2(2) + (-1)(6) -3]/sqrt(2^2 +1^2)

d = [4 - 6 - 3]/sqrt(5)

d = -5/sqrt(5) = sqrt(5) (ignoring negative sign)

Answer 3

You got started correctly, but in y = mx + b that you used, b is the y-intercept, not the y-coordinate of the point. First write the equation of the line perpendicular using
(y-y1)=m(x-x1) to give x+ 2y = 14 which you must solve simultaneously with the given equation to find the point where they intersect which is (4,5). Then find the distance from (2,6) to (4,5). You were close only by chance. I think you should get
sqrt(5).

Answer 4

shouldn't the perpendicular line be :

y - 6 = (-1/2)(x - 2)

y = -x/2 +7

distance = sqrt(5)