What is the reflection of the point (3,4) in the line 3x+4y=50?

Question

you have to go through how to do it too.

Answer 1

the reflection of point (3,4) will be on a line perpendicular to 3x+4y=50 and passing by the given point, and at equal dist from the line to that of the point(3,4) and the line.

anyway I'll try to show u how to solve that

first we have to write the equation of the line passing thru (3,4) and perpendicular of y= -3/4x + 50/3
m1 is the slope of the equation y=-3/4x + 50/3 so m1=-3/4
m2 is the slope of the equation perpendicular to the above one
m1 * m2 = -1
so m2=4/3

then we will have y=4/3x + b
replacing x and y in the above equation we get
4 = 4/3 *3 +b so b = 0
then the equation passing thru point A(3,4) and perpendicular to the equation 3x+4y=50 will be:
y = 4/3 x

Now to find the distance between point A(3,4) and the equation
y=-3/4x +50/4 we should get the intersection point of the two perpendicular lines
to do that we set the two y coordinates equal
4/3x = -3/4x + 50/4
(4/3 + 3/4)x = 50/4
25/12 x = 50/4
x = 6
and replacing x in either equation we get y=4/3 * 6 = 8
then the intersection point will be I(6,8)
Let point B(xb,yb) be the reflection of A
so I midpoint of AB
the xi = (xb + xa)/2 => 2xi - xa = xb = 2*6 - 3 = 9
and yi = (yb + ya)/2=> yb = 2yi -ya = 2*8 - 4 = 12

so the reflection of point A(3,4) is point B(9,12)

Answer 2

Let A = (3,4)
Let B = (x, y) the coord of the reflection point.

You state that the segment AB is perpendicular to line 3x + 4 y = 50, whose director vector is : (4, -3)

4(x-3) -3 (y-4) = 0

You state that the middle of AB falls on the line 3x + 4y = 50
3 (x+3) +4 ( y+4 ) = 100

You get x = 9 and y = 12

Answer 3

The point 3,4 is not on the line. Substitute x=3 in the given equation and solve for y. Next substitute computed y in the given equation and solve for x. The new computed x and y are the coordinates of the reflected point.

Answer 4

Line is 3x+4y=50 and pt is,say, A (3,4) and required is B(a,b), since B is reflection of A, so AB is perpendicular to given line.
So preduct of slopes of AB and given line should be -1
slope of given line=-3/4
slope of AB=(b-4)/a-3
so [(b-4)/a-3]* -3/4=-1 which gives 3b=4a………eq. no. (1)
Now perpendicular distance of A from given line must be equal to perpendicular distance of B from given line, in other words distance between AB should be equal to double of distance of A from given line.
perpendicular distance of A from given line= magnitude of [3(3)+4(4)-50]/sq.root of 3^2+4^2=5,
thus distance between AB=10
(a-3)^2+(b-4)^2=10^2,using (1)
(a-3)^2+(4a/3-4)^2=100,solve this to find a then put in (1) to get b