The equation of a line passing through the po | vedclass

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

Question

(A) Let the given point be $P(1, 2)$ and the external point be $Q(3, 1)$.We want to find the equation of a line passing through $P$ such that its perp

Vedclass · Accepted Answer

$y = 2x$

Vedclass · Answer

(A) Let the given point be $P(1, 2)$ and the external point be $Q(3, 1)$.We want to find the equation of a line passing through $P$ such that its perpendicular distance from $Q$ is maximized.The perpendicular distance from a point $Q$ to a line passing through $P$ is always less than or equal to the distance $PQ$ itself.The maximum distance is achieved when the line is perpendicular to the segment $PQ$ at point $P$.Let the line be $L$. Since $L \perp PQ$,the slope of $L$ is the negative reciprocal of the slope of $PQ$.Slope of $PQ = \frac{1 - 2}{3 - 1} = \frac{-1}{2}$.Therefore,the slope of line $L$ is $m = -(\frac{1}{-1/2}) = 2$.The equation of line $L$ passing through $(1, 2)$ with slope $2$ is:$y - 2 = 2(x - 1)$$y - 2 = 2x - 2$$y = 2x$.

The equation of a line passing through the point $(1, 2)$ whose distance from the point $(3, 1)$ is the greatest is

Similar Questions

The equation of the line parallel to $x/a + y/b = 1$ and passing through $(a, b)$ is .....

$A$ line passes through $(-1, -3)$ and is perpendicular to $x + 6y = 5$. Its $x$-intercept is

The equations of the lines which cut off an intercept of $-1$ from the $y$-axis and are equally inclined to the axes are

$A$ line passes through $(x_{1}, y_{1})$ and $(h, k)$. If the slope of the line is $m$,show that $k - y_{1} = m(h - x_{1})$.

The equation of the line passing through the origin and the point $(a \cos \theta, a \sin \theta)$ is .....

Vedclass Test Series

Exam Paper Generator

Online Exam Module