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Straight line

Posted November 21st, 2009 by alpha
in

One side of a rectangle lies along the line 4x+7y+5=0. Two of its vertices are (-3,1) and (1,1).Find the equations of other 3 sides.

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Solution

On November 21st, 2009 alpha says:

Let the rectangle be ABCD(taken in clockwise order)
Let side AB has eqn. 4x+7y+5=0
The coordinates of B is (-3,1).[since this point satisfies the eqn. of side AB]
(1,1) can't be the coordinate of A[it does not satisfy the eqn. of side AB]
slope of AB=-4/7
AB is perpendicular to BC.
Slope of BC=7/4
(1,1) can't be the coordinate of C [because its slope will not be equal to 7/4]
hence the coordinate of D will be (1,1).
How to proceed after this?

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

answer

On November 21st, 2009 Structure says:
$$4(x-1)-7(y-1)=0$$
$$4(x+3)-7(y-1)=0$$
$$7(x-1)+4(y-1)=0$$

How could you frame these equations? :-?

On November 24th, 2009 alpha says:

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

answer

On November 25th, 2009 Structure says:

I have just written the equations of lines by opposite points (-3,1) and (1,1) as you know the slope.

$$y-y_0=m(x-x_0)$$
$$m_1=-\frac{4}{7}$$
$$m_2=\frac{7}{4}$$

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