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