If the line passing through the point (4, -3) | vedclass

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(C) Let the slope of the required line be $m$. The slope of the line joining $(1, 1)$ and $(2, 3)$ is $m_1 = \frac{3-1}{2-1} = 2$.The angle between th

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$12$

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(C) Let the slope of the required line be $m$. The slope of the line joining $(1, 1)$ and $(2, 3)$ is $m_1 = \frac{3-1}{2-1} = 2$.The angle between the two lines is $45^{\circ}$,so $	an(45^{\circ}) = |\frac{m - m_1}{1 + m \cdot m_1}|$.$1 = |\frac{m - 2}{1 + 2m}|$.This gives two cases: $1 + 2m = m - 2$ or $1 + 2m = -(m - 2)$.Case $1$: $m = -3$. Since the slope is negative,this is a valid solution.Case $2$: $1 + 2m = -m + 2 \implies 3m = 1 \implies m = \frac{1}{3}$. This is positive,so we reject it.The equation of the line with slope $m = -3$ passing through $(4, -3)$ is $y - (-3) = -3(x - 4)$,which simplifies to $y + 3 = -3x + 12$,or $3x + y = 9$.Dividing by $9$,we get $\frac{x}{3} + \frac{y}{9} = 1$.The $x$-intercept is $a = 3$ and the $y$-intercept is $b = 9$.The sum of the intercepts is $a + b = 3 + 9 = 12$.

If the line passing through the point $(4, -3)$ and having a negative slope makes an angle of $45^{\circ}$ with the line joining the points $(1, 1)$ and $(2, 3)$,then the sum of the intercepts of that line is:

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