Answer to Question #238217 in Analytic Geometry for Pabi

Question

One side of a square lies along the straight line4𝑥 + 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

a)

Among the coordinates of the equation 4x +3y = 26 are:

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

Considering the drawing, we can solve for the vertices of the square by finding the slope of one of the vertices against the center point and apply the same to the rest.

First, with reference to the given center point (2, 3) and following the conditions that (1) one side of the square should be resting along the 4x+3y=26 line and (2) the diagonals must be intersecting at the specified point (-2,3), we can see that points (5,2) & (-1, 10) are the only possible points that can be considered for such prerequisites as vertices.

The vertices are now represented by points A, B, C, & D. And the center point as V. Let’s get the slope of the line AV as our reference.

Slope m = (Y2 - Y1) / (X2 - X1)

where

(X2, Y2) = (5,2) & (X1, Y1) = (-2, 3)

m = (2–3) / (5-(-2))

m = -1 / 7

Since we already have the coordinates for vertices A & B, we can just apply the slope for points C & D.

Point C is located at the left side of the center point

X3 = (-2 - 7) = -9

Y3 = (3-(-1)) = 4

C (X3,Y3) = (-9, 4)

Point D is at the lower part of the center point, thus switching the slope’s values .

X4 = (-2 - 1) = -3

Y4 = (3 - 7) = -4

D (X4,Y4) = (-3, -4)

The coordinates of the square’s vertices are: (5, 2), (-1, 10) (-9, 4) (-3, -4)

b)

 line 4𝑥 + 3𝑦 = 26 $\implies y=-4x/3+26/3$

slope is -4/3

for perpendicular:

slope is 3/4

then:

for AD and point (5,2):

$2=3\cdot5/4+a \implies a=-1.75$

so,

$y=3x/4-1.75$

for BC and point (-1,10):

$10=-3/4+a \implies a=10.75$

so,

$y=3x/4+10.75$