Find the y-intercept of a line that is perpen | vedclass

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

Question

(D) The given line is $3x + y = 3$,which can be written as $y = -3x + 3$. The slope of this line is $m_1 = -3$.Since the required line is perpendicula

Vedclass · Accepted Answer

$4/3$

Vedclass · Answer

(D) The given line is $3x + y = 3$,which can be written as $y = -3x + 3$. The slope of this line is $m_1 = -3$.Since the required line is perpendicular to the given line,its slope $m_2$ must satisfy $m_1 	imes m_2 = -1$.Thus,$-3 	imes m_2 = -1$,which gives $m_2 = 1/3$.The equation of the line passing through $(2, 2)$ with slope $m_2 = 1/3$ is given by $y - y_1 = m_2(x - x_1)$.Substituting the values: $y - 2 = \frac{1}{3}(x - 2)$.Multiplying by $3$: $3y - 6 = x - 2$,which simplifies to $x - 3y + 4 = 0$.To find the $y$-intercept,set $x = 0$: $0 - 3y + 4 = 0$,which gives $3y = 4$,so $y = 4/3$.Therefore,the $y$-intercept is $4/3$.

Find the $y$-intercept of a line that is perpendicular to $3x + y = 3$ and passes through the point $(2, 2)$.

Similar Questions

Let $\alpha \in R$. If the line $(\alpha+1) x+\alpha y+\alpha=1$ passes through a fixed point $(h, k)$ for all $\alpha$,then $h^2+k^2=$

If the normal form of the equation of a straight line $4x + 3y + 2 = 0$ is $x \cos \alpha + y \sin \alpha = p$ and its intercept form is $\frac{x}{a} + \frac{y}{b} = 1$,then $\frac{p \sec \alpha}{ab} = $

The equation of the line perpendicular to $2x - 3y + 5 = 0$ and making an intercept of $3$ with the positive $Y$-axis is

The equation of the line which cuts off an intercept $3$ units on $OX$ and an intercept $-2$ units on $OY$ is:

$A$ line meets the $x$-axis and $y$-axis at the points $A$ and $B$ respectively. If the midpoint of $AB$ is $(x_1, y_1)$,then the equation of the line is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module