If the lines y = 3x + 1 and 2y = x + 3 are eq | vedclass

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

Question

(D) The slopes of the given lines are $m_1 = 3$ and $m_2 = \frac{1}{2}$. Let the slope of the third line be $m_3 = m$.Since the lines are equally incl

Vedclass · Accepted Answer

$\frac{1 \pm 5\sqrt{2}}{7}$

Vedclass · Answer

(D) The slopes of the given lines are $m_1 = 3$ and $m_2 = \frac{1}{2}$. Let the slope of the third line be $m_3 = m$.Since the lines are equally inclined to the third line,the angle between the first and third line is equal to the angle between the second and third line.Using the formula for the angle between two lines,$	an 	heta = |\frac{m_a - m_b}{1 + m_a m_b}|$,we have:$|\frac{3 - m}{1 + 3m}| = |\frac{m - 1/2}{1 + m/2}|$$|\frac{3 - m}{1 + 3m}| = |\frac{2m - 1}{2 + m}|$Case $1$: $\frac{3 - m}{1 + 3m} = \frac{2m - 1}{2 + m}$$(3 - m)(2 + m) = (2m - 1)(1 + 3m)$$6 + 3m - 2m - m^2 = 2m + 6m^2 - 1 - 3m$$6 + m - m^2 = 6m^2 - m - 1$$7m^2 - 2m - 7 = 0$Using the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$m = \frac{2 \pm \sqrt{4 - 4(7)(-7)}}{14} = \frac{2 \pm \sqrt{4 + 196}}{14} = \frac{2 \pm \sqrt{200}}{14} = \frac{2 \pm 10\sqrt{2}}{14} = \frac{1 \pm 5\sqrt{2}}{7}$.

If the lines $y = 3x + 1$ and $2y = x + 3$ are equally inclined to the line $y = mx + 4$,then $m =$ ?

Similar Questions

For the lines $2x + 5y = 7$ and $2x - 5y = 9$,which of the following statements is true?

The angle between the lines $2x + 11y - 7 = 0$ and $x + 3y + 5 = 0$ is

The equations of the lines passing through $(1,1)$ and making an angle of $45^{\circ}$ with the line $x+y=0$ are

Consider the lines $L_1: x - y = 1,$ $L_2: x + y = 1,$ $L_3: 2x + 2y = 5,$ and $L_4: 2x - 2y = 7.$ Which of the following statements is true?

Find the angles between the lines $\sqrt{3}x + y = 1$ and $x + \sqrt{3}y = 1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module