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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

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

Similar Questions

The equation of the perpendicular bisector of the line segment joining $A(-2, 3)$ and $B(6, -5)$ is

What is the reflection of the point $(1, -2)$ with respect to the $y$-axis?

The image of the line $x+y-2=0$ in the $y$-axis is

The point $Q$ is the image of the point $P(1,5)$ about the line $y=x$ and $R$ is the image of the point $Q$ about the line $y=-x$. The circumcentre of the $\Delta PQR$ is

The coordinates of the foot of the perpendicular drawn from the point $(3,4)$ on the line $2x+y-7=0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module