Find the distance from the point 9,8 and the line Y=-x?

Question

hint- use distance formula, also closest distance between line and point if perpindicular. Good luck
Best answer gets 10 points,

Please show all work to get full credit

the answer is not square root 145

Answer 1

A perpendicular line provides the shortest distance from a point to a line.
Lines y = x + k are perpendicular to line y = -x.
Line y = x-1 is the line of that series that goes thru point 9,8.
The lines y=x-1 and y=-x intersect at the point where the two equations are equal.
y = x - 1
y = -x
Substitute second equation into first, gives -x = x -1
-2x = -1
x = 1/2
y = -1/2
Check. That works in both equations:
-1/2 = 1/2 -1; -1/2 = -1/2
-1/2 = -(1/2); -1/2 = -1/2
So the distance formula goes from point 9,8 to point 1/2, -1/2

distance = SQRT((9-1/2)^2 +(8--1/2)^2)
= SQRT((8.5)^2 +(8.5)^2)
= SQRT(144.25)
~ 12.02

Answer 2

The formula that computes the distance from a point (c,d) to a line a x + b y - k =0 is
D = abs(a c + b d -k)/sqrt(a^2+b^2)

Here, abs means the absolute value of the number.

So in our case, the distance D from (9,8) to y + x =0 is
D = abs(9+8)/sqrt(1^2+1^2) = 17/sqrt(2).

Answer 3

Generally: the normal to the line y=-x is y' = ax+b such that y.y' = 0 (y dot y'). But if you prefer the old school then build a right angle triangle whose vertices are the point (9,8); the incidence of the normal from this point to the line y=-x; and the intersection of the normal to the x-axis from this point to and the line y=-x at (9,-9). The normal to the line y=-x is parallel to the line y=x and therefore the angles of this isosceles triangle are 45 degrees each. Now you should be able to use symmetry and Pythagoras' theorem to figure out the length of the sides.

Answer 4

Distance:
d=√x^2+y^2
d=√(9)^2+(8)^2
d=√81+64
d=√145

Answer 5

The distance from point (9,8) to point (9,9) is 1. To point (8,8) it's also 1. The shortest distance is a perpendicular line.
If we draw a line from pt. (9,8) to (9,9), and a line from (9,8) to (8,8), and a third line from (8,8) and (9,9) we create a right triangle whose side a= 1, side b= 1 and the hypotenuse or side c= square root of 2.
Then we draw the perpendicular line from pt. (9,8) to line y=x. This line cuts side c exactly in half and gives us a right triangle with hypotenuse of 1, and side s of square root of 2 divided by 2.
Applying the Pythagorean theorem we find out that:
s^2 + distance^2 = 1
distance^2=1-s^2
since s=square root of 2 over 2
s^2=2/4=1/2
distance^2=1 - 1/2
distance = square root of 1/2
distance = 1/square root of 2

Answer 6

y=-x has a slope of -1. the distance to the line will be perpindicular to it so the slope will be 1
y=x+b
8=9+b
b=-1
y=-x
y=x-1 add
2y=-1
y=-1/2
-1/2=-x
x=1/2
check
-1/2+1/2-1
distance from (9,8) to (1/2,-1/2)
d^2=(8--1/2)^2+(9-1/2)^2
d^2=2*(8.5)2
d=8.5√2=12.02

Answer 7

Let me try it in a different approach.

Let L1 represent the segment from the origin to (9,8), and L2 represent the line y = -x. Also, let ∅ be the intersection angle between L1 and L2. Then, we have

L1 = (9,8), L2 = (1,-1)
cos ∅ = L1∙L2/|L1 L2| = -1/√290

d = |L1|sin ∅ = |L1|√(1-cos^2 ∅ ) = 12.02082

Answer 8

Distance,d = (8+9)/sqr(2)=17/sqr2 Unit.Ans.

Answer 9

d^2=(x-x1)^2+(y-y2)^2

d^2=(x-9)^2+(y-8)^2

But y=-x so substitute

d^2=(x-9)^2+(-x-8)^2

(-x-8)= - (x+8) so (-x-8)^2 = (x+8)^2 since (-1)^2=1

so d^2=(x-9)^2+(x+8)^2

d^2=(x^2-18x+81) + (x^2+16x+64)=2x^2-2x+145

x = -b/(2a) is the minimum for a parabola that opens up and this occurs at
x=2/(2*2) = 1/2 = 0.5, so since y=-x, y=-0.5

Now all that's left is to find the distance between (9,8) and (0.5,-0.5).

d^2=(x-9)^2+(-x-8)^2 = (0.5-9)^2+(-0.5-8)^2=144.5

So d=12.02 approx.

Answer 10

sqrt145