Let a variable line of slope m 0 passing th | vedclass

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

Question

(A) The equation of the line passing through $(4, -9)$ with slope $m$ is given by $y + 9 = m(x - 4)$.To find the $x$-intercept $A$,set $y = 0$: $9 = m

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) The equation of the line passing through $(4, -9)$ with slope $m$ is given by $y + 9 = m(x - 4)$.To find the $x$-intercept $A$,set $y = 0$: $9 = m(x - 4) \Rightarrow x = 4 + \frac{9}{m}$. Thus,$A = (4 + \frac{9}{m}, 0)$.To find the $y$-intercept $B$,set $x = 0$: $y + 9 = m(-4) \Rightarrow y = -9 - 4m$. Since $B$ is a point on the axis and we consider distance,we look at the magnitude. Given $m > 0$,the $y$-coordinate is negative,so $B = (0, -(9 + 4m))$.The sum of the distances from the origin is $S = OA + OB = |4 + \frac{9}{m}| + |-(9 + 4m)|$.Since $m > 0$,$S = 4 + \frac{9}{m} + 9 + 4m = 13 + 4m + \frac{9}{m}$.Using the Arithmetic Mean-Geometric Mean inequality $(AM \geq GM)$: $\frac{4m + \frac{9}{m}}{2} \geq \sqrt{4m 	imes \frac{9}{m}} = \sqrt{36} = 6$.Therefore,$4m + \frac{9}{m} \geq 12$.Thus,$S \geq 13 + 12 = 25$.

Let a variable line of slope $m > 0$ passing through the point $(4, -9)$ intersect the coordinate axes at the points $A$ and $B$. The minimum value of the sum of the distances of $A$ and $B$ from the origin is

Similar Questions

The number of integer values of $m$,for which the $x$-coordinate of the point of intersection of the lines $3x + 4y = 9$ and $y = mx + 1$ is also an integer,is

$P$ 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 coordinates of the vertices $A$ and $B$ of an isosceles triangle $ABC$ $(AC = BC)$ are $(-2, 3)$ and $(2, 0)$ respectively. $A$ line parallel to $AB$ and having a $y$-intercept equal to $\frac{43}{12}$ passes through $C$. Then the coordinates of $C$ are:

The point on the line $3x + 4y = 5$ which is equidistant from $(1, 2)$ and $(3, 4)$ is

Consider the fourteen lines in the plane given by $y=x+r$ and $y=-x+r$,where $r \in \{0, 1, 2, 3, 4, 5, 6\}$. The number of squares formed by these lines,whose sides are of length $\sqrt{2}$,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module