Let PQR be a right angled isosceles triangle, right angled at P (2, 1). If the equation of the line

Q: Wait, why is the angle $45^\circ$? I thought it was $90^\circ$?

Great question! The $90^\circ$ angle is at vertex $P$, between the legs $PQ$ and $PR$. But the triangle is isosceles, meaning the angles at the base ($Q$ and $R$) must be equal. Since the sum of angles in a triangle is $180^\circ$ and the right angle takes up $90^\circ$, the remaining $90^\circ$ must be split equally between the other two angles. Thus, each base angle is $45^\circ$.

Q: Why do we get two slopes from the absolute value?

The absolute value $\left| \dots \right|$ accounts for the two possible directions of the lines relative to the hypotenuse. One slope corresponds to $PQ$ and the other to $PR$. Both are valid and necessary to define the pair of lines!

Q: How do we combine the equations at the end?

If you have two lines $L_1 = 0$ and $L_2 = 0$, their combined equation is simply $L_1 \cdot L_2 = 0$. This represents the locus of all points lying on either line. By multiplying the two linear equations we found, we get the joint equation for the pair.

Question

Let $PQR$ be a right angled isosceles triangle, right angled at $P (2, 1)$. If the equation of the line $QR$ is $2x + y = 3$, then the equation representing the pair of lines $PQ$ and $PR$ is

Options:

(A) $3x^2 - 3y^2 + 8xy + 20x + 10y + 25 = 0$

(B) $3x^2 - 3y^2 + 8xy - 20x - 10y + 25 = 0$

(C) $3x^2 - 3y^2 + 8xy + 10x + 15y + 20 = 0$

(D) $3x^2 - 3y^2 - 8xy - 10x - 15y - 20 = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a coordinate geometry problem; we are uncovering the hidden symmetry of a triangle.

Imagine standing at the vertex $P(2, 1)$, looking out at the hypotenuse $QR$. The problem states that $	riangle PQR$ is a right-angled isosceles triangle.

In an isosceles right triangle, the angles at the base are equal. Since the sum of angles in a triangle is $180^{\circ}$ and the right angle takes up $90^{\circ}$, the remaining $90^{\circ}$ must be split equally between the other two angles.

Thus, each base angle is $45^{\circ}$. This means the lines $PQ$ and $PR$ are both inclined at $45^{\circ}$ to the hypotenuse $QR$. This is the geometric soul of the problem.

### The Slope Hunt

Now, let us find the slope of $QR$. The equation is $2x + y = 3$. Rearranging this into the slope-intercept form $y = mx + c$, we get $y = -2x + 3$.

The slope $m_{QR}$ is $-2$. We need the slopes of $PQ$ and $PR$. Let these slopes be $m$.

We use the angle formula:

$$	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$$

With $	heta = 45^{\circ}$, we know $	an 45^{\circ} = 1$. Substituting our values, we get:

$$1 = \left| \frac{m - (-2)}{1 + m(-2)} ight|$$

This simplifies to:

$$1 = \left| \frac{m + 2}{1 - 2m} ight|$$

This absolute value represents the two distinct paths the lines $PQ$ and $PR$ take from vertex $P$.

### The Algebraic Dance

Solving for the positive case, $1 = \frac{m + 2}{1 - 2m}$, we find $1 - 2m = m + 2$, which leads to $3m = -1$, or $m = -\frac{1}{3}$.

Solving for the negative case, $-1 = \frac{m + 2}{1 - 2m}$, we find $-1 + 2m = m + 2$, which gives $m = 3$. We have our slopes.

Now, we use the point-slope form $y - y_1 = m(x - x_1)$ with $P(2, 1)$.

For $m = -\frac{1}{3}$, the equation is $y - 1 = -\frac{1}{3}(x - 2)$, which simplifies to $x + 3y - 5 = 0$.

For $m = 3$, the equation is $y - 1 = 3(x - 2)$, which simplifies to $3x - y - 5 = 0$.

### The Grand Finale

Finally, to find the combined equation, we multiply these two linear equations:

$$(x + 3y - 5)(3x - y - 5) = 0$$

Expanding this product, we distribute the terms:

$$x(3x - y - 5) + 3y(3x - y - 5) - 5(3x - y - 5) = 0$$

This yields:

$$3x^2 - xy - 5x + 9xy - 3y^2 - 15y - 15x + 5y + 25 = 0$$

Grouping like terms, we arrive at the final result:

**$$3x^2 - 3y^2 + 8xy - 20x - 10y + 25 = 0$$**

This is the elegant solution to our problem. Geometry is truly a language of patterns, and once you see them, the math flows effortlessly.

Animated Solution for Mathematics - Straight Lines: Let $P QR$ be a right angled isosceles triangle, right angled at $P (2, 1)$ . If the equation of the line $QR$ is $2 x + y = 3$ , then the equation representing the pair of lines $P Q$ and $P R$ is

Select Answer:

Visualized Solution

Visualizing the Triangle $P QR$

Geometry of Isosceles Right Triangle

Finding the Slope of $QR$

Applying the Angle Formula

Substituting Values

Solving for Slopes (Case 1)

Solving for Slopes (Case 2)

Setting up Line Equations

Equation of Line $P Q$

Equation of Line $P R$

Forming the Combined Equation

Expanding the Product

Final Simplification

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

The Slope Hunt

The Algebraic Dance

The Grand Finale

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