Asked by Lilly
Sorry I'm posting so many. I'm on a schedule and panicking.
Which is the distance between the point with the coordinates (1,2) and the line with the equation 2x-3y=-2?
Which is the distance between the point with the coordinates (1,2) and the line with the equation 2x-3y=-2?
Answers
Answered by
Steve
distance from (h,k) to ax+by+c=0 is
|ah+bk+c|/√(a^2+b^2)
so, you have
|1*2 - 2*3 + 2|/√(4+9) = 2/√13
or, if you can't recall the formula, you can work it out by realizing that the shortest distance will be along a perpendicular line through the point.
2x-3y = -2 has slope 2/3
The perpendicular thus has slope -3/2
Now you have a point and a slope, so the equation of the perpendicular line through (1,2) is
y-2 = -3/2 (x-1)
y = -3/2 x + 7/2
The perpendicular intersects at (17/13,20/13)
The distance from (1,2) to (17/13,20/13) is √(16/169+36/169) = √52/13 = 2/√13
|ah+bk+c|/√(a^2+b^2)
so, you have
|1*2 - 2*3 + 2|/√(4+9) = 2/√13
or, if you can't recall the formula, you can work it out by realizing that the shortest distance will be along a perpendicular line through the point.
2x-3y = -2 has slope 2/3
The perpendicular thus has slope -3/2
Now you have a point and a slope, so the equation of the perpendicular line through (1,2) is
y-2 = -3/2 (x-1)
y = -3/2 x + 7/2
The perpendicular intersects at (17/13,20/13)
The distance from (1,2) to (17/13,20/13) is √(16/169+36/169) = √52/13 = 2/√13
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