Find the equation of the line passing through | vedclass

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

Question

(D) Let the points be $A(-5, 6)$ and $B(-6, 5)$.The slope of the line $AB$ is given by $m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 6}{-6 - (-5)} =

Vedclass · Accepted Answer

$x + y - 5 = 0$

Vedclass · Answer

(D) Let the points be $A(-5, 6)$ and $B(-6, 5)$.The slope of the line $AB$ is given by $m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 6}{-6 - (-5)} = \frac{-1}{-1} = 1$.Since the required line is perpendicular to $AB$,its slope $m_2$ must satisfy $m_1 	imes m_2 = -1$.Thus,$1 	imes m_2 = -1$,which gives $m_2 = -1$.The equation of a line with slope $m$ passing through $(x_0, y_0)$ is given by $y - y_0 = m(x - x_0)$.Substituting $m = -1$ and the point $(2, 3)$,we get:$y - 3 = -1(x - 2)$$y - 3 = -x + 2$$x + y - 5 = 0$.

Find the equation of the line passing through the point $(2, 3)$ and perpendicular to the line joining the points $(-5, 6)$ and $(-6, 5)$.

Similar Questions

If $(a, 8)$ is a point on the line segment joining $(2, 5)$ and $(4, -1)$,then the value of $a$ is:

Given that lines $L_1: y=m_a x, L_2: y=m_b x$ and $L_3: y=m_c x$ make equal intercepts on the line $x+y=1$,then

The equation of a line passing through the origin and perpendicular to the line joining $(a, 0)$ and $(-a, 0)$ is

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

The straight line $L$ is perpendicular to the line $5x - y = 1$. If the area of the triangle formed by the line $L$ and the coordinate axes is $5$,find the equation of the line $L$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module