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

Q: Wait, how do I know if the given line is a side or a diagonal?

Excellent question! In coordinate geometry, always test your points. If a point satisfies the equation, it lies on the line. If the line were a diagonal, neither vertex would necessarily lie on it in this specific way. By testing $A(-3, 1)$, we confirmed it lies on the line, proving the line is a side.

Q: Why do we swap coefficients and change the sign for perpendicular lines?

Think about the slope. If a line has slope $m = -A/B$, a perpendicular line must have slope $m' = -1/m = B/A$. The equation $Bx - Ay + k = 0$ naturally gives a slope of $B/A$. It's a beautiful, direct consequence of the negative reciprocal slope property!

Q: Is there a faster way to find the constants?

While you could use distance formulas, the 'family of lines' approach we used is the most elegant and least prone to error. It keeps your work clean and logical, which is the hallmark of a JEE topper.

Question

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 the other three sides.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at a rectangle. You have been given one side, defined by the equation:

$$4x + 7y + 5 = 0$$

You are also provided with two vertices, $A(-3, 1)$ and $C(1, 1)$. The first step in any JEE Advanced problem is to orient yourself by determining if the given line represents a side or a diagonal.

By substituting point $A(-3, 1)$ into the line equation, we find:

$$4(-3) + 7(1) + 5 = -12 + 7 + 5 = 0$$

It fits perfectly! This confirms that the line is indeed a side of the rectangle.

However, when we test point $C(1, 1)$, we get:

$$4(1) + 7(1) + 5 = 16 
eq 0$$

Since this is not zero, point $C$ is not on this line. This confirms that $C$ must be the vertex diagonally opposite to $A$.

### The Parallel Universe

In a rectangle, opposite sides are parallel. This is a powerful property because parallel lines share the same slope.

If our side $AB$ is $4x + 7y + 5 = 0$, any line parallel to it must take the form:

$$4x + 7y + k = 0$$

We know the side $CD$ passes through $C(1, 1)$. By substituting these coordinates into our general form, we get:

$$4(1) + 7(1) + k = 0 \Rightarrow 11 + k = 0 \Rightarrow k = -11$$

Thus, the equation of the opposite side $CD$ is:

$$4x + 7y - 11 = 0$$

### The Perpendicular Shift

Now, we turn our attention to the adjacent sides. In a rectangle, adjacent sides are perpendicular.

For a line $Ax + By + C = 0$, a perpendicular line is given by $Bx - Ay + k' = 0$. Applying this to our original line $4x + 7y + 5 = 0$, we get the form:

$$7x - 4y + k' = 0$$

Since this side $BC$ must pass through $C(1, 1)$, we substitute the values:

$$7(1) - 4(1) + k' = 0 \Rightarrow 3 + k' = 0 \Rightarrow k' = -3$$

The equation for side $BC$ is:

$$7x - 4y - 3 = 0$$

### Final Calculation

Finally, we need the last side, $AD$. This side is parallel to $BC$, so it must have the same form:

$$7x - 4y + k'' = 0$$

Since it passes through $A(-3, 1)$, we substitute:

$$7(-3) - 4(1) + k'' = 0 \Rightarrow -21 - 4 + k'' = 0 \Rightarrow k'' = 25$$

The equation for side $AD$ is:

$$7x - 4y + 25 = 0$$

We have successfully mapped out the entire rectangle. This process is fundamentally about understanding the rigid, beautiful structure of Euclidean geometry.

Animated Solution for Mathematics - Straight Lines: One side of a rectangle lies along the line $4 x + 7 y + 5 = 0$ . Two of its vertices are $(- 3, 1)$ and $(1, 1)$ . Find the equations of the other three sides.

Visualized Solution

$V i s u a l i z in g t h e G i v e n I n f or ma t i o n$

$C h ec k in g V er t e x A (- 3, 1)$

$C h ec k in g V er t e x C (1, 1)$

$E q u a t i o n o f O pp os i t e S i d e C D$

$F in d in g C o n s t an t k f or S i d e C D$

$E q u a t i o n o f S i d e C D$

$E q u a t i o n o f A d j a ce n t S i d e B C$

$F in d in g C o n s t an t k^{'} f or S i d e B C$

$E q u a t i o n o f S i d e B C$

$E q u a t i o n o f A d j a ce n t S i d e A D$

$F in d in g C o n s t an t k^{''} f or S i d e A D$

$E q u a t i o n o f S i d e A D$

$F ina l E q u a t i o n s o f t h e R ec t an g l e^{'} s S i d es$

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Parallel Universe

The Perpendicular Shift

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Given Information

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Checking Vertex A(−3,1)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Checking Vertex C(1,1)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of Opposite Side CD

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding Constant k for Side CD

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of Side CD

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of Adjacent Side BC

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding Constant k′ for Side BC

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of Side BC

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of Adjacent Side AD

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding Constant k′′ for Side AD

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of Side AD

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Equations of the Rectangle′s Sides