If two distinct points lying on the line x+y= | vedclass

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

Question

(A) Let the point on the line $x+y=4$ be $(\alpha, 4-\alpha)$.Given that the perpendicular distance from this point to the line $4x+3y-10=0$ is $1$.Us

Vedclass · Accepted Answer

$10\sqrt{2}$

Vedclass · Answer

(A) Let the point on the line $x+y=4$ be $(\alpha, 4-\alpha)$.Given that the perpendicular distance from this point to the line $4x+3y-10=0$ is $1$.Using the distance formula: $\left|\frac{4\alpha+3(4-\alpha)-10}{\sqrt{4^2+3^2}}\right|=1$.$\left|\frac{4\alpha+12-3\alpha-10}{5}\right|=1$.$|\alpha+2|=5$.This gives $\alpha+2=5$ or $\alpha+2=-5$.So,$\alpha=3$ or $\alpha=-7$.For $\alpha=3$,the point is $(3, 1)$.For $\alpha=-7$,the point is $(-7, 11)$.The distance $d$ between $(3, 1)$ and $(-7, 11)$ is given by $\sqrt{(-7-3)^2+(11-1)^2}$.$d=\sqrt{(-10)^2+(10)^2} = \sqrt{100+100} = \sqrt{200} = 10\sqrt{2}$.

If two distinct points lying on the line $x+y=4$ are at a unit distance from the line $4x+3y-10=0$,and the distance between these two points is $d$,then the value of $d$ is:

Similar Questions

Reduce the equation $x-y=4$ into the normal form $x \cos \omega + y \sin \omega = p$. Find the perpendicular distance from the origin $(p)$ and the angle between the perpendicular and the positive $x$-axis $(\omega)$.

What is the distance between the parallel lines $y = 2x + 4$ and $6x = 3y + 5$?

The equation of the base of an equilateral triangle is $12x+5y-65=0$. If one of its vertices is $(2,3)$,then the length of the side is

Let the line $L$ drawn perpendicular to the lines $2x - 3y + 4 = 0$ and $6x - 9y + 7 = 0$ meet them at $A$ and $B$ respectively. If $P(1, 1)$ is a point on $L$,then the ratio in which $P$ divides $AB$ is

What is the length of the perpendicular from the point $(3, 4)$ to the line $3x + 4y + 10 = 0$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module