Angle Between Two Lines

1. Find the original vertex $$V$$

Let the coordinates of $$V$$ be $$(x_1, y_1)$$. Its reflection across the line $$2x + 3y - 5 = 0$$ is given as $$V'(-1, -2)$$.

Using the standard line reflection formula: $$\frac{x_1 - x_2}{a} = \frac{y_1 - y_2}{b} = -2 \frac{ax_2 + by_2 + c}{a^2 + b^2}$$

Substituting our values ($$x_2 = -1$$, $$y_2 = -2$$, $$a = 2$$, $$b = 3$$, $$c = -5$$):

$$\frac{x_1 - (-1)}{2} = \frac{y_1 - (-2)}{3} = -2 \frac{2(-1) + 3(-2) - 5}{2^2 + 3^2}$$

$$\frac{x_1 + 1}{2} = \frac{y_1 + 2}{3} = -2 \left(\frac{-2 - 6 - 5}{13}\right) = -2 \left(\frac{-13}{13}\right) = 2$$

Now, solve for $$x_1$$ and $$y_1$$:

$$\frac{x_1 + 1}{2} = 2 \implies x_1 + 1 = 4 \implies x_1 = 3$$

$$\frac{y_1 + 2}{3} = 2 \implies y_1 + 2 = 6 \implies y_1 = 4$$

Thus, the vertex opposite to the hypotenuse is $$V(3, 4)$$.

2. Find the slopes of the perpendicular sides

The given hypotenuse line is $$4x + 3y = 5$$, which has a slope of $$m = -\frac{4}{3}$$.

Since the triangle is a right-angled isosceles triangle, both perpendicular sides intersect at $$V$$ and make an angle of $$45^\circ$$ with the hypotenuse. Let $$m_s$$ be the slope of a side:

$$\tan(45^\circ) = \left\vert \frac{m_s - \left(-\frac{4}{3}\right)}{1 + m_s\left(-\frac{4}{3}\right)} \right\vert \implies 1 = \left\vert \frac{3m_s + 4}{3 - 4m_s} \right\vert$$

This gives two cases:

Case 1:

$$3m_s + 4 = 3 - 4m_s \implies 7m_s = -1 \implies m_s = -\frac{1}{7}$$

Case 2:

$$3m_s + 4 = -(3 - 4m_s) \implies 3m_s + 4 = -3 + 4m_s \implies m_s = 7$$

(Note that the product of the two slopes is $$(-\frac{1}{7}) \times 7 = -1$$, confirming they are perpendicular).

3. Write the equations of the sides

Both lines pass through the calculated vertex $$V(3, 4)$$: 

Side 1 (with slope $$m_s = 7$$): $$y - 4 = 7(x - 3) \implies y - 4 = 7x - 21 \implies 7x - y = 17$$

Side 2 (with slope $$m_s = -\frac{1}{7}$$): $$y - 4 = -\frac{1}{7}(x - 3) \implies 7y - 28 = -x + 3 \implies x + 7y = 31$$

Correct Answer is  (C) $$7x - y = 17$$ and $$x + 7y = 31$$

The family of lines is given by $$L_1 + \lambda L_2 = 0$$, where:

• $$L_1: 3x + 2y - 10 = 0$$
• $$L_2: x - 2y = 0$$

The fixed point $$P$$ is the intersection of $$L_1$$ and $$L_2$$. Substituting $$x = 2y$$ from $$L_2$$ into $$L_1$$:

$$3(2y) + 2y - 10 = 0 \implies 8y = 10 \implies y = \frac{5}{4}$$

Substituting $$y$$ back into $$x = 2y$$:

$$x = 2\left(\frac{5}{4}\right) = \frac{5}{2}$$

Thus, the fixed point is $$P = \left(\frac{5}{2}, \frac{5}{4}\right)$$.

The required line makes an angle of $$45^\circ$$ with the line $$L: 5x - y + 4 = 0$$.

The slope of line $$L$$ is $$m_L = 5$$. Let the slope of the required line be $$m$$. Using the tangent formula:

$$\tan(45^\circ) = \left| \frac{m - 5}{1 + 5m} \right|$$ $$1 = \left| \frac{m - 5}{1 + 5m} \right|$$

Case I:

$$\frac{m - 5}{1 + 5m} = 1 \implies m - 5 = 1 + 5m \implies -4m = 6 \implies m = -\frac{3}{2}$$

Case II:

$$\frac{m - 5}{1 + 5m} = -1 \implies m - 5 = -1 - 5m \implies 6m = 4 \implies m = \frac{2}{3}$$

Based on the provided solution logic, we select Case I where $$m = -\frac{3}{2}$$. This slope corresponds to the member where $$\lambda = 0$$:

$$3x + 2y - 10 = 0$$

To find the intercepts:

• $$x$$-intercept ($$a$$): Set $$y = 0$$ $$3x = 10 \implies a = \frac{10}{3}$$
• $$y$$-intercept ($$b$$): Set $$x = 0$$ $$2y = 10 \implies b = 5$$

The question asks for the sum of the absolute values of the $$x$$-intercept and $$y$$-intercept:

$$\text{Sum} = |a| + |b|$$ $$\text{Sum} = \left| \frac{10}{3} \right| + |5|$$ $$\text{Sum} = \frac{10}{3} + \frac{15}{3}$$ $$\text{Sum} = \frac{25}{3}$$

Let $$L_1: 3x+y-7=0$$ and the angle bisector be $$B: x-2y+5=0.$$

First, we find the point of intersection $$P$$ of the line $$L_1$$ and the bisector $$B$$. From the bisector equation, we can express $$x$$ as:

$$x = 2y - 5$$

Substituting this into the equation for $$L_1$$:

$$3(2y - 5) + y - 7 = 0 \implies 7y - 22 = 0 \implies y = \frac{22}{7}$$

Substituting $$y$$ back to find $$x$$:

$$x = 2\left(\frac{22}{7}\right) - 5 = \frac{9}{7}$$

Thus, both lines must intersect at the point $$P\left(\frac{9}{7}, \frac{22}{7}\right)$$.

Next, let the angle between $$L_1$$ and $$B$$ be $$\theta$$. The slopes are $$m_1 = -3$$ and $$m_B = \frac{1}{2}$$. Using the angle formula:

$$\tan\theta = \left|\frac{m_1 - m_B}{1 + m_1 m_B}\right| = \left|\frac{-3 - \frac{1}{2}}{1 + (-3)\left(\frac{1}{2}\right)}\right| = \left|\frac{-\frac{7}{2}}{-\frac{1}{2}}\right| = 7$$

Let the slope of the required line $$L_2$$ be $$m$$. Since $$B$$ is the angle bisector, the angle between $$L_2$$ and $$B$$ must also be $$\theta$$:

$$\left|\frac{m - \frac{1}{2}}{1 + \frac{m}{2}}\right| = 7 \implies \frac{2m - 1}{2 + m} = \pm 7$$

To find the line on the opposite side of the bisector, we solve the negative branch:

$$\frac{2m - 1}{2 + m} = -7 \implies 2m - 1 = -14 - 7m$$

$$9m = -13 \implies m = -\frac{13}{9}$$

Now, using the point-slope form for the line $$L_2$$ passing through $$P\left(\frac{9}{7}, \frac{22}{7}\right)$$ with slope $$m = -\frac{13}{9}$$:

$$y - \frac{22}{7} = -\frac{13}{9}\left(x - \frac{9}{7}\right)$$

Multiplying the entire equation by 63 to clear the denominators:

$$63y - 198 = -13(7x - 9)$$

$$63y - 198 = -91x + 117$$

$$91x + 63y - 315 = 0$$

Dividing by the common factor of 7 yields the final line equation:

$$13x + 9y - 45 = 0$$