The equation of a given straight line is frac | vedclass

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

Question

(C) The given equation of the line is $\frac{x-x_1}{\cos 	heta}=\frac{y-y_1}{\sin 	heta}=\gamma$. This represents a line passing through $(x_1, y_1)

Vedclass · Accepted Answer

$\cot 	heta$

Vedclass · Answer

(C) The given equation of the line is $\frac{x-x_1}{\cos 	heta}=\frac{y-y_1}{\sin 	heta}=\gamma$. This represents a line passing through $(x_1, y_1)$ with an inclination $	heta$. The slope of this line is $m_1 = 	an 	heta$. Let the slope of the line perpendicular to this line be $m_2$. Since the lines are perpendicular,$m_1 	imes m_2 = -1$. Therefore,$m_2 = -\frac{1}{	an 	heta} = -\cot 	heta$. The equation of the required line is given as $\frac{x}{a}+\frac{y}{b}=1$,which can be rewritten as $y = -\frac{b}{a}x + b$. The slope of this line is $m_2 = -\frac{b}{a}$. Equating the slopes,we get $-\frac{b}{a} = -\cot 	heta$. Thus,$\frac{b}{a} = \cot 	heta$.

The equation of a given straight line is $\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}=\gamma$. If the equation of the line perpendicular to the given line and passing through $(\alpha, \beta)$ is $\frac{x}{a}+\frac{y}{b}=1$,then $\frac{b}{a}$ is equal to

Similar Questions

Suppose a line makes an angle of $120^{\circ}$ with the positive direction of $X$-axis. If the length of the perpendicular from the origin to that line is $4$,then the equation of the line is

If a line $L$ passing through the point $A(-2, 4)$ makes an angle of $60^{\circ}$ with the positive direction of $X$-axis in the anti-clockwise direction and $B(p, q)$ lying in the $3^{\text{rd}}$ quadrant is a point on $L$ at a distance of $6$ units from the point $A$,then $\sqrt{p^2+q^2-8q} = $

If the three points $(3q, 0)$, $(0, 3p)$, and $(1, 1)$ are collinear, then which one is true?

Find the slope of the line passing through the points $(3, -2)$ and $(7, -2)$.

$A$ line passing through the point $P(1, 1)$ and parallel to the line $x - y = 5$ cuts the line $x + 3y - 2 = 0$ at $Q$. Then twice the length of the segment $PQ$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module