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 $$x + 7y = 31$$
91- 6303239042
