i need help with a math problem. Find the distance from point 6,10 to the line Y=-X?

Answer 1

The formula for the distance between a line
ax + by + c = 0

and a point (h,k)
is

|ah + bk + c|/√(a² + b²)

Let's put the equation of the line into proper form.
y = -x
x + y = 0

Also (h,k) = (6,10)

Now we can apply the formula.

|ah + bk + c|/√(a² + b²)
= |(1)(6) + (1)(10) + 0|/√(1² + 1²)
= |6 + 10|/√(1 + 1)
= 16/√2 = 16(√2)/2 = 8√2

The distance between the point and the line is 8√2.

Answer 2

Find the distance from point 6,10 to the line Y=-X?
The line y=-x has a slope of -1 so a line perpendicular to this line must have a slope of 1.

The equation of the line is y=x+b
10= 6 + b so b= 4
The equation is therefore y=x+4
So the lines intersect at x=-2 and y = 2
Thus the distance is sqrt(6-(-2))^2+ (10-2)^2)= sqrt(64+64)
= 8sqrt(2)

Answer 3

line A: y = -x <=> x + y = 0 ( in form of Ax + By + C = 0 with A = B = 1 and C = 0)
M(6,10)
d(M,A) (distance from point M to line A)
= abs(A*x_M + B*y_M + C) / sqrt(A^2 + B^2) ( dash here means subscript)
= abs(6 + 10) / sqrt( 1^2 + 1^2) = 16 / sqrt(2) = 11.31

Answer 4

this perpindicular from (6, 10) to th eline y=-x has a slope of 1

y=x+b
10=6+b
b=4
y=x+4
y=-x add
2y=4
y=2
2=-x
x=-2

the distance from (6, 10) to (-2, 2) is
d=√((10-2)^2+(6--2)^2
d=√(8^2+8^2=8√2=11.314