A is the point of intersection of the lines 3 | vedclass

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

Question

(B) Step $1$: Find the intersection point $A$ of the lines $3x + y - 4 = 0$ and $x - y = 0$. Adding the equations,$4x = 4 \Rightarrow x = 1$. Substitu

Vedclass · Accepted Answer

$x + 2y = 3$

Vedclass · Answer

(B) Step $1$: Find the intersection point $A$ of the lines $3x + y - 4 = 0$ and $x - y = 0$. Adding the equations,$4x = 4 \Rightarrow x = 1$. Substituting $x = 1$ in $x - y = 0$,we get $y = 1$. Thus,$A = (1, 1)$.Step $2$: Let the slope of the required line be $m$. The slope of the line $x - 3y + 5 = 0$ is $m_1 = \frac{1}{3}$.Step $3$: The angle between the lines is $45^{\circ}$. Using the formula $	an 	heta = \left| \frac{m - m_1}{1 + m \cdot m_1} ight|$,we have $	an 45^{\circ} = \left| \frac{m - 1/3}{1 + m/3} ight| = 1$.Step $4$: This gives $\frac{m - 1/3}{1 + m/3} = 1$ or $\frac{m - 1/3}{1 + m/3} = -1$.Case $1$: $m - 1/3 = 1 + m/3$ $\Rightarrow \frac{2m}{3} = \frac{4}{3}$ $\Rightarrow m = 2$.Case $2$: $m - 1/3 = -1 - m/3$ $\Rightarrow \frac{4m}{3} = -2/3$ $\Rightarrow m = -1/2$.Since the slope is negative,we choose $m = -1/2$.Step $5$: The equation of the line passing through $(1, 1)$ with slope $m = -1/2$ is $y - 1 = -1/2(x - 1)$ $\Rightarrow 2y - 2 = -x + 1$ $\Rightarrow x + 2y = 3$.

$A$ is the point of intersection of the lines $3x + y - 4 = 0$ and $x - y = 0$. If a line having a negative slope makes an angle of $45^{\circ}$ with the line $x - 3y + 5 = 0$ and passes through $A$,then its equation is:

Similar Questions

For a point $P(x, y)$ in the plane,let $d_1(P)$ and $d_2(P)$ be the distances of the point $P$ from the lines $x-y=0$ and $x+y=0$ respectively. The area of the region $R$ consisting of all points $P$ lying in the first quadrant of the plane and satisfying $2 \leq d_1(P)+d_2(P) \leq 4$ is:

$A$ line is such that its segment between the lines $5x - y + 4 = 0$ and $3x + 4y - 4 = 0$ is bisected at the point $(1, 5)$. Obtain its equation.

The points $(1, 1)$,$(0, \sec^2 \theta)$,and $(\csc^2 \theta, 0)$ are collinear for:

$A$ pair of straight lines drawn through the origin form an isosceles right-angled triangle with the line $2x + 3y = 6$. Find the equations of the lines and the area of the triangle thus formed.

The distance of the point $(1, 2)$ from the line $x + y = 0$ measured parallel to the line $3x - y = 2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module