Q.Find the equation of one of the sides of an isosceles right angled triangle whose hypotenuse is given by $3x+4y=4$ and the opposite vertex of the hypotenuse is $(2,2)$.

The key idea here is using the angle between two lines in an isosceles right-angled triangle.

1. The slope of the hypotenuse $3x+4y=4$ is $m_h = -3/4$.
2. In an isosceles right-angled triangle, the two equal sides make an angle of $45^\circ$ with the hypotenuse. Let $m$ be the slope of one of these sides.
3. Using the formula for the angle between two lines, $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$, we have: $\tan 45^\circ = 1 = \left| \frac{m - (-3/4)}{1 + m(-3/4)} \right| = \left| \frac{4m+3}{4-3m} \right|$.
4. This yields two possible slopes for the sides:
   • $\frac{4m+3}{4-3m} = 1 \implies 4m+3 = 4-3m \implies 7m=1 \implies m=1/7$. …

Let's break down this problem by first understanding the geometry of an isosceles right-angled triangle.

An isosceles right-angled triangle has two equal sides (called legs) and one right angle ($90^\circ$). The side opposite the right angle is the hypotenuse. Because it's isosceles, the two angles opposite the equal sides must also be equal. Since the sum of angles in a triangle is $180^\circ$, and one angle is $90^\circ$, the other two angles must each be $45^\circ$.

In this problem, we are given:

1. The equation of the hypotenuse: $3x+4y=4$.
2. The coordinates of the vertex opposite the hypotenuse: $(2,2)$. This vertex is where the two equal sides (legs) meet, meaning it is the right-angled vertex of the triangle. Let's call this vertex $A(2,2)$.

The key insight here is that the two legs of the triangle pass through vertex $A(2,2)$, and each of these legs makes an angle of $45^\circ$ with the hypotenuse. This property allows us to find the slopes of the legs.

Here's how we proceed:

1. Determine the slope of the hypotenuse.

   The equation of the hypotenuse is $3x+4y=4$. To find its slope, we can rewrite it in the slope-intercept form $y=mx+c$.

   $4y = -3x + 4$

   $y = -\frac{3}{4}x + 1$

   The slope of the hypotenuse, $m_H$, is $-\frac{3}{4}$.

2. Formulate the equation of a line passing through the right-angled vertex.

   Let $m$ be the slope of one of the legs. Since this leg passes through $A(2,2)$, its equation can be written using the point-slope form:

   $y - y_1 = m(x - x_1)$

   $y - 2 = m(x - 2)$

3. Use the angle condition between the leg and the hypotenuse.

   We know that the angle between each leg and the hypotenuse is $45^\circ$. We can use the formula for the angle $\theta$ between two lines with slopes $m_1$ and $m_2$:

   $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$

   Here, $\theta = 45^\circ$, $m_1 = m$ (slope of the leg), and $m_2 = m_H = -\frac{3}{4}$.

   Since $\tan 45^\circ = 1$, we have:

   $1 = \left| \frac{m - (-\frac{3}{4})}{1 + m(-\frac{3}{4})} \right|$

   $1 = \left| \frac{m + \frac{3}{4}}{1 - \frac{3}{4}m} \right|$

   To simplify the fraction inside the absolute value, multiply the numerator and denominator by 4:

   $1 = \left| \frac{4m + 3}{4 - 3m} \right|$

4. Solve for the possible slopes ($m$) of the legs.

   The absolute value equation gives us two possibilities: …