The equation of the straight line in the norm | vedclass

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

Question

(B) Let the required line be $x+2y+k=0$. Since it divides the distance between $x+2y+3=0$ and $x+2y+8=0$ in the ratio $1:2$ internally,the constant $k

Vedclass · Accepted Answer

$x \cos \alpha+y \sin \alpha=\frac{14}{\sqrt{45}}, \alpha=\pi+	an ^{-1} 2$

Vedclass · Answer

(B) Let the required line be $x+2y+k=0$. Since it divides the distance between $x+2y+3=0$ and $x+2y+8=0$ in the ratio $1:2$ internally,the constant $k$ satisfies $\frac{|k-3|}{|k-8|} = \frac{1}{2}$. $2|k-3| = |k-8| \Rightarrow 4(k^2-6k+9) = k^2-16k+64$. $3k^2-8k-28=0 \Rightarrow (3k-14)(k+2)=0$. Since the line lies between the two given lines,$k$ must be between $3$ and $8$,so $k = \frac{14}{3}$. The equation is $x+2y+\frac{14}{3}=0$,or $3x+6y+14=0$. To convert to normal form $x \cos \alpha + y \sin \alpha = p$,we write $-3x-6y=14$. Dividing by $\sqrt{(-3)^2+(-6)^2} = \sqrt{45}$,we get $\frac{-3}{\sqrt{45}}x + \frac{-6}{\sqrt{45}}y = \frac{14}{\sqrt{45}}$. Here $\cos \alpha = \frac{-3}{\sqrt{45}} = \frac{-1}{\sqrt{5}}$ and $\sin \alpha = \frac{-6}{\sqrt{45}} = \frac{-2}{\sqrt{5}}$. Since both $\cos \alpha$ and $\sin \alpha$ are negative,$\alpha$ lies in the third quadrant. $\alpha = \pi + 	an^{-1}(\frac{-2/-1}) = \pi + 	an^{-1}(2)$.

The equation of the straight line in the normal form which is parallel to the lines $x+2y+3=0$ and $x+2y+8=0$ and divides the distance between these two lines in the ratio $1:2$ internally is

Similar Questions

If ${x_1}, {x_2}, {x_3}$ and ${y_1}, {y_2}, {y_3}$ are both in $G$.$P$. with the same common ratio,then the points $({x_1}, {y_1}), ({x_2}, {y_2})$ and $({x_3}, {y_3})$:

$P$ is a point on either of the two lines $y - \sqrt{3}|x| = 2$ at a distance of $5 \ units$ from their point of intersection. The coordinates of the foot of the perpendicular from $P$ on the bisector of the angle between them are

The point on the line $4x - y - 2 = 0$ which is equidistant from the points $(-5, 6)$ and $(3, 2)$ is:

$A$ line is such that its segment between the straight lines $5x - y - 4 = 0$ and $3x + 4y - 4 = 0$ is bisected at the point $(1, 5)$,then its equation is

$A$ line $4x + y = 1$ passes through the point $A(2, -7)$ and meets the line $BC$,whose equation is $3x - 4y + 1 = 0$,at the point $B$. The equation of the line $AC$ such that $AB = AC$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module