Find the distance between Line L and Point P if Line L contains the points
1 answer:
Answer:
(b) 4.2
Step-by-step explanation:
The distance from a point to a line can be found using the formula. To use it, we need the equation of the line in general form.
We notice that the coordinates of the given points have the same y/x = 2/3 ratio. This means they lie on the line y = 2/3x. The equation for that line, written in general form is ...
y = 2/3x
3y = 2x
2x -3y = 0 . . . . . . general form equation for the line through the points.
The formula for the distance from (x, y) to the line ax+by+c=0 is ...
d = |ax +by +c|/√(a²+b²)
The distance from (x, y) to our line is given by the formula ...
d = |2x -3y|/√(2² +(-3)²)
d = |2x -3y|/√13
For point P(9, 11), the distance to the line is ...
d = |2·9 -3·11|/√13 = |18 -33|/√13 = 15/√13 ≈ 4.2
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Given:
The triangle ABC is reflected over the x-axis.
The distance between point A and A’ is 14.
To find:
The distance between the x-axis and A’.
Solution:
If a figure is reflected across the x-axis then the corresponding parts are mirror image of each other about the x-axis.
It means the distance between A and x-axis is same as the distance between x-axis and A'.
The distance between point A and A’ is 14.
Let d be the distance between the x-axis and A’. Then,
Therefore, the correct option is 1.