The number of lines which pass through the po | vedclass

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

Question

(D) Let the point $P = (2, -3)$ and the point $Q = (-1, 2)$.Any line passing through $P$ will have a distance from $Q$ that is at most the distance $P

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Let the point $P = (2, -3)$ and the point $Q = (-1, 2)$.Any line passing through $P$ will have a distance from $Q$ that is at most the distance $PQ$.The distance $PQ = \sqrt{(2 - (-1))^2 + (-3 - 2)^2} = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$.Since $\sqrt{34} \approx 5.83$, and we are looking for a line at a distance of $8$ from $Q$, we note that $8 > \sqrt{34}$.Because the perpendicular distance from a point to a line cannot exceed the distance between the point and a fixed point on the line, no such line exists.Thus, the number of such lines is $0$.

The number of lines which pass through the point $(2, -3)$ and are at a distance of $8$ from the point $(-1, 2)$ is:

Similar Questions

Identify the point on the line $2x + 3y + 7 = 0$,which is at a distance of $3$ units from $(1, -3)$.

The equations of two lines passing through $(0, a)$ which are at a distance of $a$ from the point $(2a, 2a)$ are:

If $O$ is the origin and $P, Q$ are points on the line $3x + 4y + 15 = 0$ such that $OP = OQ = 9$, then the area of $\triangle OPQ$ is (in $\sqrt{2}$)

The distance between the lines $3x + 4y = 9$ and $6x + 8y = 15$ is equal to units.

Statement-$1$: There is one line through $A(4, -5)$ such that its distance from $B(-2, 3)$ is $12$.
Statement-$2$: $AB = 10$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module