Given the four lines with equations x + 2y = | vedclass

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

Question

(D) The equations of the lines are:$L_1: x + 2y - 3 = 0$$L_2: 3x + 4y - 7 = 0$$L_3: 2x + 3y - 4 = 0$$L_4: 4x + 5y - 6 = 0$First,check for parallel lin

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) The equations of the lines are:$L_1: x + 2y - 3 = 0$$L_2: 3x + 4y - 7 = 0$$L_3: 2x + 3y - 4 = 0$$L_4: 4x + 5y - 6 = 0$First,check for parallel lines by comparing slopes $(m = -A/B)$:$m_1 = -1/2, m_2 = -3/4, m_3 = -2/3, m_4 = -4/5$.Since no slopes are equal,no two lines are parallel. Thus,they cannot form a rectangle.Next,check for concurrency for any three lines. For $L_1, L_2, L_3$:$\Delta = \begin{vmatrix} 1 & 2 & -3 \ 3 & 4 & -7 \ 2 & 3 & -4 \end{vmatrix} = 1(-16 + 21) - 2(-12 + 14) - 3(9 - 8) = 5 - 4 - 3 = -2 
eq 0$.Since the determinant is not zero,the lines are not concurrent.Therefore,the correct option is $D$.

Given the four lines with equations $x + 2y = 3,$ $3x + 4y = 7,$ $2x + 3y = 4,$ and $4x + 5y = 6,$ these lines are:

Similar Questions

For all values of $a$ and $b$,the line $(a+2b)x + (a-b)y + (a+5b) = 0$ passes through a fixed point. Find that point.

The points $A(1, 2)$,$B(2, 4)$,and $C(4, 8)$ form a/an

If $x+2y-3=0$,$3x+4y-7=0$,$2x+3y-4=0$,and $4x+5y-6=0$ are the equations of four lines,then

Find the equation of the line passing through the intersection of the lines $4x - 3y - 1 = 0$ and $5x - 2y - 3 = 0$ and parallel to the line $2x - 3y + 2 = 0$.

$(x_1, y_1)$ is the point of concurrency of a family of lines. If the algebraic sum of the lengths of the perpendiculars drawn to these lines from $(2, 0)$,$(0, 2)$,and $(1, 1)$ is zero,then $(x_1, y_1) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module