Let the lines be L_1: 2x + 3y - 7 = 0 and L_2 | vedclass

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

Question

(D) The given lines are $L_1: 2x + 3y - 7 = 0$ and $L_2: 2x + 3y - 12 = 0$.For point $A(3, -5)$,evaluate the expressions:$L_1(3, -5) = 2(3) + 3(-5) -

Vedclass · Accepted Answer

None of these.

Vedclass · Answer

(D) The given lines are $L_1: 2x + 3y - 7 = 0$ and $L_2: 2x + 3y - 12 = 0$.For point $A(3, -5)$,evaluate the expressions:$L_1(3, -5) = 2(3) + 3(-5) - 7 = 6 - 15 - 7 = -16 $L_2(3, -5) = 2(3) + 3(-5) - 12 = 6 - 15 - 12 = -21 Since both values have the same sign,point $A$ does not lie between the lines.Perpendicular distance $d_1$ from $A$ to $L_1$ is $\frac{|2(3) + 3(-5) - 7|}{\sqrt{2^2 + 3^2}} = \frac{|-16|}{\sqrt{13}} = \frac{16}{\sqrt{13}}$.Perpendicular distance $d_2$ from $A$ to $L_2$ is $\frac{|2(3) + 3(-5) - 12|}{\sqrt{2^2 + 3^2}} = \frac{|-21|}{\sqrt{13}} = \frac{21}{\sqrt{13}}$.The distance between the parallel lines is $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} = \frac{|-7 - (-12)|}{\sqrt{2^2 + 3^2}} = \frac{5}{\sqrt{13}}$.Thus,none of the given options $A, B, C$ are correct.

Let the lines be $L_1: 2x + 3y - 7 = 0$ and $L_2: 2x + 3y - 12 = 0$. For the point $A(3, -5)$,which of the following is true?

Similar Questions

Find the distance between the parallel lines $15x + 8y - 34 = 0$ and $15x + 8y + 31 = 0$.

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

The base of an equilateral triangle is along the line given by $3x + 4y = 9$. If a vertex of the triangle is $(1, 2)$,then the length of a side of the triangle is

If the equation of a line parallel to $3x - 2y + 5 = 0$ and at a distance of $5$ units from it is $3x - 2y + C = 0$,then $C$ is equal to

If two sides of a square are $4x + 3y - 20 = 0$ and $4x + 3y + 15 = 0$,then the area of the square is

Vedclass Test Series

Exam Paper Generator

Online Exam Module