Grade 10 Math Help?

Question

Distance from a point to a line?

(-1, -3) and x+3y-9=0

It must be the shortest distance from point (-1,-3) to the line x+3y-9=0

Answer 1

x+3y-9 = 0 can be rearranged to get:

y = (-x/3) + 3

No to work out the distance you need to see what line is perpendicular to this one and this will have a gradient 3

y = 3x + c

Finally you need to work out the length of this line between the point (-1,-3) and the intercept with your original:

y = 3x

To find the intercept both equations must be equal

3x = (-x/3) + 3
9x = -x + 9
10x = 9
x = 0.9

By placing into either equation you get
y = 2.7

Now use Pythagoras theorem to find the distance between

(-1,-3) and (0.9, 2.7)

Difference in x = 1.9
Difference in y = 5.7

Distance between points = SQRT (1.9^2 + 5.7^2)

SQRT (3.61 + 32.49)

= 6.00832 or 6.0 (2 sig fig)

Answer 2

The is an equation of a line called "normal equation"
it is(ax+by+c)(/(+-sqrt(a^2+b^2))=0 which has the property that when you substytute (x,y) by the coordinates of any point
the value which takes the left side is the distance of the point to the line.
The explanation is longer than the workout
(x+3y-9)/(+-sqrt(10))=0so
d= (-1-9-9)(+-sqrt(10)) = taken positive =19/sqtr10= (19sqrt10)/10

Answer 3

by defination, the distance from a point to a line is dertermind by the distance of a segment than runs perpendicular from that point to the line.

first, you have to find the line that intersect x + 3y - 9 = 0 and is perpendicular to that line

put the equation into slope form
x + 3y - 9 = 0
3y = -x + 9
y = -1/3x + 3

the slope that is perpendicular to that line is 3
y = mx + b
-3 = 3(-1) + b
-3 = -3 + b
b = 0
y = 3x

now find the point where the line intercept.
plug 3x for y
x + 3(3x) - 9 = 0
x + 9x = 9
10x = 9
x = .9

y = 3(9/10)
y = 2.7

you have 2 points. (-1,-3) and (.9 , 2.7)
use distance fomular

d^2 = (y2 - y1)^2 + (x2 - x1)^2

d^2 = (2.7 - (-3) )^2 + (.9 - (-1) )^2

d^2 = 32.49 + 3.61

d^2 = 36.1

d = sqrt(36.1) or about 6

Answer 4

Let the point on the line that results in the shorted distance from (-1, -3) be ((9-3a), a) ---> [from the equation x+3y-9=0.]

Gradient of line:
x+3y-9=0
3y=-x+9
y = (-1/3)x + 3
Gradient of line is (-1/3).

For two line to be perpendicular, m1 x m2 = -1,
m1 x (-1/3) = -1
m1 = 3

So gradient of the perpendicular line is 3.
(-1, -3), ((9-3a), a)

(-3-a)/[-1-(9-3a)] = 3
(-3-a)/(-10+3a) = 3
-3 - a = 3(-10 + 3a)
-3 - a = -30 + 9a
30 - 3 = 9a + a
27 = 10a
a = 2.7
So 9-3a = 9-3(2.7) = 0.9

The point is (0.9,2.7).

Distance
= SQRT[(0.9-(-1))^2 + (2.7-(-3))^2]
= SQRT[3.61 + 32.49]
= SQRT36.1
= 6.01