4. the diagram at right shows the graph of 3x + 4y = 12. the shaded figure is a square, three of whose vertices are on the coordinate axes. the fourth vertex is on the line. find (a) the x- and y-intercepts of the line; (b) the length of a side of the square. (c) show that your point is equidistant from the coordinate axes.

Remark

This is a very interesting question.  Draw a line from the origin to where the upper right vertex of the square touches the line. That line has the property that the its equation is y = x. So the "solution" to the point of intersection is the solution of the two equations.

y =  x            (1)

3x + 4y = 12 (2)

Put x in for y in equation 2

3x + 4x = 12

7x = 12

x = 12/7

x = 1.714

y = 1.714

Problem A

x intercept

The x intercept occurs when y = 0

3x + 4(0) = 12

3x = 12                Divide by 3

x = 12/3

x = 4

the x intercept = (4,0)

y intercept

The y intercept occurs when x =0

3(0) + 4y = 12

4y = 12

y = 12/4

y = 3

y intercept = (0,3)

Problem B

x and y both equal 1.714 so they are also the length of the square's side.

Problem C

See solution above. x =y is the key fact.

x = y = 1.714

8(-10x+3.5y-7)

8×-10x + 8×3.5y + 8×-7

-80x + 28y - 56

therefore

-80x+28y-56 is the correct answer

step-by-step explanation: