The coordinates of a point on the line x+y+1= | vedclass

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

Question

(B, D) Let $(h, k)$ be a point on the line $x+y+1=0.$ Since the point lies on the line, we have $h+k+1=0,$ which implies $h = -k-1.$ The perpendicular

Vedclass · Accepted Answer

$(-1, 0)$

Vedclass · Answer

(B, D) Let $(h, k)$ be a point on the line $x+y+1=0.$ Since the point lies on the line, we have $h+k+1=0,$ which implies $h = -k-1.$ The perpendicular distance from $(h, k)$ to the line $3x+4y+2=0$ is given by $\frac{|3h+4k+2|}{\sqrt{3^2+4^2}} = \frac{1}{5}.$ Substituting $h = -k-1$ into the distance formula: $\frac{|3(-k-1)+4k+2|}{5} = \frac{1}{5}$ $|-3k-3+4k+2| = 1$ $|k-1| = 1$ This gives two cases: $k-1 = 1 \implies k=2,$ then $h = -2-1 = -3.$ Point is $(-3, 2).$ $k-1 = -1 \implies k=0,$ then $h = 0-1 = -1.$ Point is $(-1, 0).$ Thus, the required points are $(-3, 2)$ and $(-1, 0).$ Both options $B$ and $D$ are correct.

The coordinates of a point on the line $x+y+1=0,$ which is at a distance of $\frac{1}{5}$ unit from the line $3x+4y+2=0,$ are

Similar Questions

For the three points $A(2,0)$,$B(0,2)$,and $P(1,1)$,suppose $d$ is the algebraic sum of the distances of $A$ and $B$ from a line that passes through $P$. Then,which of the following is correct?

$A$ line has slope $m$ and $y$-intercept $4$. The distance between the origin and the line is equal to

$A$ straight line makes equal negative intercepts on the coordinate axes. If the perpendicular distance from the origin to the line is $4 \text{ units}$, then the equation of the line is:

If the line $3x + 4y + \lambda = 0$ divides the distance between the lines $3x + 4y + 5 = 0$ and $3x + 4y - 5 = 0$ in the ratio of $3:7$,then a value of $\lambda$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module