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Question: What is the reflection of the point?

What is the reflection of the point (3,4) in the line 3x+4y=50? How do I work it out?

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Answer:

reflection across 3x + 4y = 50?

this will be equidistant along the perpendicular
4y = -3x + 50
y = -3/4 x + 12.5

slope is -3/4; thus, the perpendicular line will have slope 4/3

the perpendicular line will have slope 4/3 and contain (3 , 4)
y - 4 = 4/3 (x - 3) ==> y = 4/3 x

at the intersection with the given line:
4/3 x = -3/4 x + 12.5
(4/3 + 3/4) x = 12.5
(16/12 + 9/12) x = 25/2
x = 6
and when x = 6, then y = 8
intersection point of perpendicular line ==> (6 , 8)

this will be midpoint of segment from (3 , 4) to the reflection

in x, it is 3 from the point to the midpoint, so add 3 more to the reflection
in y, it is 4 from the point to the midpoint, so add 4 more to the reflection

reflection will be (9 , 12)

First, convert the line to slope intercept form. It yields more info easier, and is simply easier to use.

4y = -3x + 50

y = -3/4 x + 50/4

The reflected point will be equidistant from the line on the other side of the line from the original point.

Distances from lines are ALWAYS perpendiculars to the lines.

Then we need the slope of the perpendicular line through the point.

Perpendicular lines have slopes that are negative inverses of one another, so the slope of the perpendicular must be 4/3

Find the equation of the perpendicular that goes through the point.

y = 4/3 x + b

4 = 4/3 * 3 + b

4 = 4 + b so b = 0

Then the eq of the perpendicular is y=4/3 x

Now find the point of intersection between the perpendicular and the line.
Notice here that they both equal y, so set them equal to each other.

4/3 x = -3/4 x + 50/4

(16/12 + 9/12)x = 25/2

25/12 x = 25/2

x = 6, then y = 4/3 *6 = 8, so the point of intersection is (6,8)

The distance from (3,4) to (6,8) is sqrt[3^2 + 4^2] = 5

Now we need another point that is on the perpendicular (y=4/3 x) and 5 units away from point (6,8)

Let that point be (x,y)

Then the distance formula 5^2 = (x-6)^2 + (y-8)^2

But y = 4/3 x, so

5^2 = (x - 6)^2 + (4/3 x - 8)^2 {now multiply this out}

25 = x^2 - 12x + 36 + 16/9 x^2 - 64/3 x + 64

0 = x^2 - 12x + 36 + 16/9 x^2 - 64/3 x + 64 - 25 {simplify and combine like terms}

0 = 25/9 x^2 - 100/3 x + 75

Now use the quadratic formula

x = [100/3 (+/-) sqrt[10000/9 - 4 * 25/9 * 75]] / (50/9)

x = [100/3 (+/-) sqrt[10000/9 - 7500/9]] / (50/9)

x = [100/3 (+/-) sqrt[2500/9]] / (50/9)

x = [100/3 (+/-) 50/3 ]/ (50/9)

x = 150/3 / (50/9) OR 50/3 / (50/9)

x = 9 OR 3

3 is a result we should have expected because that is the x coordinate of the original point. If that isn't one of our solutions here we have done something wrong, so that is a good number to see.

So the other is 9.

Then the equation of the perpendicular is y=4/3*x, so y = 4/3 * 9 = 12

The point reflected is (9, 12)

Check out the diagram.
EG is the original line, DF is the perpendicular, F is (3,4), E = (6,8), and D = (9,12)

I drew the circle to show that the points were equidistant from Point E, and therefore the original line.

I hope this helps