Question

One side of a square lies along the straight line 4x+3y = 26. The diagonals of the square intersect at the point (-2,3).
Find;
a. The coordinates of the vertices of the square.
b. The equation of the sides of the square which are perpendicular to the given line.

Answer 1

Given: One side of a square lies along the straight line 4x+3y = 26, diagonals of the square intersect at the point (-2,3).

To find: The coordinates of the vertices of the square.

Solution:

• Now by putting value of x and y, we can determine some of the line’s coordinates because the question says that one of the sides of the square lies along this line.

• So some coordinates are:

              (-4, 14), (-1, 10), ( 2, 6), (5, 2), (8, -2), (11, 6)

• Now, with reference to the center of the diagonal (2, 3) and the conditions given in the question we can see that points (5,2) & (-1, 10) are the possible points that would be considered as vertices.

• Let the vertices represented be A, B, C, D and centre point be V.

• We know that Slope m = (Y2 - Y1) / (X2 - X1)

• Slope of AV will be:

             m = (2–3) / (5-(-2))

             m = -1 / 7

             where (X2, Y2) = (5,2) & (X1, Y1) = (-2, 3)

• For point C,  using X1, Y1

             X3 = (-2 - 7) = -9

             Y3 = (3-(-1)) = 4

             C (X3,Y3) = (-9, 4)

• and similarly for point D, using X1, Y1

            X4 = (-2 - 1) = -3

            Y4 = (3 - 7) = -4

            D (X4,Y4) = (-3, -4)

Answer:

So, the coordinates of the square vertices are (5, 2), (-1, 10) (-9, 4) (-3, -4)

Answer 2

Answer:

wala kang respeto wag niu na guys sagutin mga questions niya kase nag sasabi siya ng bad words doon sa sagot nung isa eh kung ikw yung murahin magagalit ka rin naman at the same time