If the family of straight lines ax + by + c = | vedclass

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(D) The given equation of the family of lines is $ax + by + c = 0$. Given the condition $2a + 3b = 4c$,we can rewrite this as $\frac{2a}{4} + \frac{3b

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$\left(\frac{-5}{8}, \frac{-3}{8}\right)$

Vedclass · Answer

(D) The given equation of the family of lines is $ax + by + c = 0$. Given the condition $2a + 3b = 4c$,we can rewrite this as $\frac{2a}{4} + \frac{3b}{4} = c$,or $a(\frac{1}{2}) + b(\frac{3}{4}) + c = 0$. Comparing this with $ax + by + c = 0$,we find the point of concurrency $P$ is $(\frac{1}{2}, \frac{3}{4})$. Wait,re-evaluating the condition: $2a + 3b - 4c = 0 \Rightarrow a(\frac{2}{4}) + b(\frac{3}{4}) + c(-1) = 0$. Thus,the point $P$ is $(\frac{1}{2}, \frac{3}{4})$. Let the foot of the perpendicular from $P(\frac{1}{2}, \frac{3}{4})$ to the line $x + y + 1 = 0$ be $(h, k)$. Using the formula $\frac{h - x_1}{a} = \frac{k - y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2}$: $\frac{h - 1/2}{1} = \frac{k - 3/4}{1} = -\frac{1/2 + 3/4 + 1}{1^2 + 1^2} = -\frac{9/4}{2} = -\frac{9}{8}$. $h = \frac{1}{2} - \frac{9}{8} = \frac{4 - 9}{8} = -\frac{5}{8}$. $k = \frac{3}{4} - \frac{9}{8} = \frac{6 - 9}{8} = -\frac{3}{8}$. Thus,the foot of the perpendicular is $(-\frac{5}{8}, -\frac{3}{8})$.

If the family of straight lines $ax + by + c = 0$,where $2a + 3b = 4c$,is concurrent at the point $P(l, m)$,then the foot of the perpendicular drawn from $P$ to the line $x + y + 1 = 0$ is

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