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Let l be the line given by

3x - 4y = 18,
and let P = (-6, 1). There are two squares which can be constructed such that one
side of the square is on the line l and P is one of the vertices of the square. Find
the vertices of one of these squares.

User Lee H
by
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1 Answer

3 votes

Answer:

Explanation:

To construct a square with one side on the line l, we need to find the equation of a line perpendicular to l that passes through P. The slope of the line l is 3/4, so the slope of any line perpendicular to l is -4/3. Thus, the equation of the line through P with slope -4/3 is

y - 1 = (-4/3)(x + 6)

Simplifying, we get

y = (-4/3)x - 7

To find the intersection of this line with l, we can substitute y = (-4/3)x - 7 into the equation for l:

3x - 4((-4/3)x - 7) = 18

Solving for x, we get x = -24/5. Substituting back, we get y = -13/5.

So the intersection point of the two lines is (-24/5, -13/5), which we'll call Q.

To construct a square with side PQ, we need to find the midpoint M of PQ, and then find the points R and S that are equidistant from M and P.

The midpoint of PQ is

(((-24/5) + (-6))/2, ((-13/5) + 1)/2) = (-15/5, -6/5) = (-3, -1/5)

The distance between P and M is

sqrt((-3 - (-6))^2 + (-1/5 - 1)^2) = sqrt(45.04) ≈ 6.71

So the distance between M and R (or S) is also 6.71.

Since P has coordinates (-6, 1), and the distance from M to P is 6.71, we know that R and S must have coordinates (-6 ± 6.71, 1 ± 6.71), which simplifies to (-12.71, 7.71) and (0.71, -5.71), respectively.

So one possible set of vertices for the square is PQRSA, where P = (-6, 1), Q = (-24/5, -13/5), R = (-12.71, 7.71), S = (0.71, -5.71).

User Thenickname
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