The equations of the lines passing through (1 | vedclass

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

Question

(D) The given line is $x+y=0$,which can be written as $y = -x$. The slope of this line is $m_1 = -1$.Let the slope of the required lines be $m$. The a

Vedclass · Accepted Answer

$x-1=0, y-1=0$

Vedclass · Answer

(D) The given line is $x+y=0$,which can be written as $y = -x$. The slope of this line is $m_1 = -1$.Let the slope of the required lines be $m$. The angle between the lines is $45^{\circ}$.Using the formula $	an 	heta = \left| \frac{m - m_1}{1 + m m_1} ight|$,we have:$	an 45^{\circ} = \left| \frac{m - (-1)}{1 + m(-1)} ight|$$1 = \left| \frac{m+1}{1-m} ight|$This gives two cases:Case $1$: $\frac{m+1}{1-m} = 1 \implies m+1 = 1-m \implies 2m = 0 \implies m = 0$.The equation of the line passing through $(1,1)$ with slope $0$ is $y-1 = 0(x-1) \implies y-1 = 0$.Case $2$: $\frac{m+1}{1-m} = -1 \implies m+1 = -1+m \implies 1 = -1$,which is impossible (this implies the line is vertical).$A$ vertical line passing through $(1,1)$ is $x=1$,or $x-1=0$.Thus,the equations of the lines are $x-1=0$ and $y-1=0$.

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

Similar Questions

If the lines $y=3x+1$ and $2y=x+3$ are equally inclined to the line $y=mx+4$,find the value of $m$.

The equations of the lines passing through the origin and making an angle of $60^{\circ}$ with the line $x + y\sqrt{3} + 3\sqrt{3} = 0$ are

The acute angle between the lines $x-3=0$ and $x+y=19$ is: (in $^{\circ}$)

The lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ are perpendicular to each other,if

If $3x + 4y - 5 = 0$ and $4x + ky - 8 = 0$ are two parallel lines,find the value of $k$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module