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Animated Solution for Mathematics - Straight Lines: The straight lines $x + y = 0, 3 x + y - 4 = 0, x + 3 y - 4 = 0$ form a triangle which is

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Visualized Solution

Visualizing the Three Lines

Given lines:
Line 1 ( $L_{1}$ ): $x + y = 0$
Line 2 ( $L_{2}$ ): $3 x + y - 4 = 0$
Line 3 ( $L_{3}$ ): $x + 3 y - 4 = 0$

Strategy: Finding the Vertices

To find the type of triangle, we need to find the coordinates of its vertices.
Vertices are found by solving the equations of the lines pairwise:
1. Intersection of $L_{1}$ and $L_{3}$ (Vertex $A$ )
2. Intersection of $L_{1}$ and $L_{2}$ (Vertex $B$ )
3. Intersection of $L_{2}$ and $L_{3}$ (Vertex $C$ )

Solving $L_{1}$ and $L_{3}$ for Vertex $A$

From $L_{1}$ : $y = - x$
Substitute $y = - x$ into $L_{3}$ :
$x + 3 (- x) - 4 = 0$
$- 2 x - 4 = 0 ⟹ x = - 2$
Since $y = - x$ , $y = 2$
Vertex $A = (- 2, 2)$

Solving $L_{1}$ and $L_{2}$ for Vertex $B$

From $L_{1}$ : $y = - x$
Substitute $y = - x$ into $L_{2}$ :
$3 x + (- x) - 4 = 0$
$2 x - 4 = 0 ⟹ x = 2$
Since $y = - x$ , $y = - 2$
Vertex $B = (2, - 2)$

Solving $L_{2}$ and $L_{3}$ for Vertex $C$

From $L_{2}$ : $y = 4 - 3 x$
Substitute into $L_{3}$ :
$x + 3 (4 - 3 x) - 4 = 0$
$x + 12 - 9 x - 4 = 0$
$- 8 x + 8 = 0 ⟹ x = 1$
Substitute $x = 1$ into $y = 4 - 3 x ⟹ y = 1$
Vertex $C = (1, 1)$

The Formed Triangle $A B C$

We have the three vertices:
Vertex $A = (- 2, 2)$
Vertex $B = (2, - 2)$
Vertex $C = (1, 1)$
These vertices enclose the triangle $A B C$ .

The Distance Formula

To classify the triangle, we calculate the lengths of its sides.
Distance Formula:
$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$

Calculating Side $A B$

Between $A (- 2, 2)$ and $B (2, - 2)$ :
$A B = (2 - (- 2))^{2} + (- 2 - 2)^{2}$
$A B = 4^{2} + (- 4)^{2}$
$A B = 16 + 16 = 32 = 42$

Calculating Side $B C$

Between $B (2, - 2)$ and $C (1, 1)$ :
$B C = (1 - 2)^{2} + (1 - (- 2))^{2}$
$B C = (- 1)^{2} + 3^{2}$
$B C = 1 + 9 = 10$

Calculating Side $C A$

Between $C (1, 1)$ and $A (- 2, 2)$ :
$C A = (- 2 - 1)^{2} + (2 - 1)^{2}$
$C A = (- 3)^{2} + 1^{2}$
$C A = 9 + 1 = 10$

Conclusion: Isosceles Triangle

We have:
$B C = C A = 10$
$A B = 42$
Since two sides are equal ( $B C = C A$ ), the triangle is isosceles.

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The Sigma Insight: Various Forms of Equations of a Line

Solution Diagram

Analyzing the Setup

Imagine you are standing on a vast coordinate plane. You have three lines:

L_{1} : x + y = 0

L_{2} : 3 x + y - 4 = 0

L_{3} : x + 3 y - 4 = 0

These lines are not just abstract equations; they are paths that carve out a triangular region in space. Our mission is to uncover the nature of this triangle.

The Hunt for Vertices

To understand the triangle, we must first identify its corners—the vertices. These are the points where our lines meet, found by solving the equations pairwise.

For Vertex

A

, the intersection of

L_{1}

and

L_{3}

, we use the fact that

L_{1}

gives us

y = - x

. Substituting this into

L_{3}

x + 3 (- x) - 4 = 0

- 2 x = 4 \Rightarrow x = - 2, y = 2

Thus,

A = (- 2, 2)

For Vertex

B

, the intersection of

L_{1}

and

L_{2}

, we substitute

y = - x

into

L_{2}

3 x + (- x) - 4 = 0

2 x = 4 \Rightarrow x = 2, y = - 2

Thus,

B = (2, - 2)

For Vertex

C

, the intersection of

L_{2}

and

L_{3}

, we solve the system:

3 x + y = 4

x + 3 y = 4

Subtracting the equations or using substitution yields

x = 1

and

y = 1

. So,

C = (1, 1)

Measuring the Sides

Now that we have our vertices

A (- 2, 2)

B (2, - 2)

, and

C (1, 1)

, we measure the sides using the distance formula:

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

For side

A B

A B = (2 - (- 2))^{2} + (- 2 - 2)^{2} = 4^{2} + (- 4)^{2} = 32 = 42

For side

B C

B C = (1 - 2)^{2} + (1 - (- 2))^{2} = (- 1)^{2} + 3^{2} = 10

For side

C A

C A = (- 2 - 1)^{2} + (2 - 1)^{2} = (- 3)^{2} + 1^{2} = 10

The Verdict

Look at our results:

B C = 10

and

C A = 10

Because two sides are equal, this is the hallmark of an isosceles triangle. The third side,

A B = 42

, is different, confirming that it is not equilateral.

We have successfully used coordinate geometry to prove the nature of our triangle. It is a beautiful, symmetric, isosceles triangle.

Common Traps & FAQs

Wait, why do we need to find the vertices? Can't we just look at the slopes of the lines?

That is a brilliant intuition! While looking at the slopes can tell you if lines are perpendicular (if the product of slopes is

- 1

), it doesn't directly tell you the side lengths. Since we need to classify the triangle (isosceles, equilateral, etc.), calculating the side lengths using the vertices is the most robust and foolproof method.

Is there a faster way to find the vertices without solving all three pairs of equations?

You could potentially use the properties of the lines' symmetry, but for a triangle with three distinct lines, solving the systems of equations is actually quite fast and minimizes the risk of conceptual errors. It is the standard, reliable path in coordinate geometry.

What if the lines were parallel? Would they still form a triangle?

Great question! If any two lines were parallel, they would never intersect, and thus they would not form a triangle. In this problem, the slopes are

- 1

- 3

, and

- 1/3

. Since no two slopes are equal, the lines are not parallel and will indeed intersect to form a triangle.

Animated Solution for Mathematics - Straight Lines: The straight lines $x + y = 0, 3 x + y - 4 = 0, x + 3 y - 4 = 0$ form a triangle which is

Select Answer:

Visualized Solution

Visualizing the Three Lines

Strategy: Finding the Vertices

Solving $L_{1}$ and $L_{3}$ for Vertex $A$

Solving $L_{1}$ and $L_{2}$ for Vertex $B$

Solving $L_{2}$ and $L_{3}$ for Vertex $C$

The Formed Triangle $A B C$

The Distance Formula

Calculating Side $A B$

Calculating Side $B C$

Calculating Side $C A$

Conclusion: Isosceles Triangle

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Hunt for Vertices

Measuring the Sides

The Verdict

Similar Questions

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List-I

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Animated Solution for Mathematics - Straight Lines: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The straight lines x+y=0,3x+y−4=0,x+3y−4=0 form a triangle which is

Select Answer:

Visualized Solution

Visualizing the Three Lines

Strategy: Finding the Vertices

The Formed Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABC

The Distance Formula

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

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

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

Conclusion: Isosceles Triangle

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Hunt for Vertices

Measuring the Sides

The Verdict

Similar Questions

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Animated Solution for Mathematics - Straight Lines: The straight lines $x + y = 0, 3 x + y - 4 = 0, x + 3 y - 4 = 0$ form a triangle which is

The Formed Triangle $A B C$

Calculating Side $A B$

Calculating Side $B C$

Calculating Side $C A$

A straight line through the point $A (3, 4)$ is such that its intercept between the axes is bisected at $A$ . Its equation is

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If a straight line passing through the point $P (- 3, 4)$ is such that its intercepted portion between the coordinate axes is bisected at $P$ , then its equation is :

Let $P = (- 1, 0), Q = (0, 0)$ and $R = (3, 33)$ be three points. The equation of the bisector of the angle $P QR$ is

Let $P = (- 1, 0), Q = (0, 0)$ and $R = (3, 33)$ be three points. Then the equation of the bisector of the angle $P QR$ is

If one of the lines given by $6 x^{2} - x y + 4 c y^{2} = 0$ is $3 x + 4 y = 0$ , then $c$ equals