The intercept form of the equation of the str | vedclass

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

Question

(B) Let the points be $A(4, -3)$,$B(1, 1)$,and $C(2, 3)$.The slope of line $BC$ is $m_{BC} = \frac{3-1}{2-1} = \frac{2}{1} = 2$.Since the required lin

Vedclass · Accepted Answer

$\frac{x}{-2} + \frac{y}{-1} = 1$

Vedclass · Answer

(B) Let the points be $A(4, -3)$,$B(1, 1)$,and $C(2, 3)$.The slope of line $BC$ is $m_{BC} = \frac{3-1}{2-1} = \frac{2}{1} = 2$.Since the required line is perpendicular to $BC$,its slope $m$ is given by $m = -\frac{1}{m_{BC}} = -\frac{1}{2}$.The equation of the line passing through $(4, -3)$ with slope $m = -\frac{1}{2}$ is given by the point-slope form:$y - y_1 = m(x - x_1)$$y - (-3) = -\frac{1}{2}(x - 4)$$y + 3 = -\frac{1}{2}(x - 4)$$2y + 6 = -x + 4$$x + 2y = -2$To convert this to intercept form $\frac{x}{a} + \frac{y}{b} = 1$,divide both sides by $-2$:$\frac{x}{-2} + \frac{2y}{-2} = \frac{-2}{-2}$$\frac{x}{-2} + \frac{y}{-1} = 1$.Thus,option $B$ is correct.

The intercept form of the equation of the straight line passing through the point $(4, -3)$ and perpendicular to the line passing through the points $(1, 1)$ and $(2, 3)$ is

Similar Questions

Find the slope of the line passing through the points $(3, -2)$ and $(3, 4)$.

The line joining the points $(-1, 3)$ and $(4, -2)$ will pass through the point $(p, q)$ if

The line passing through $\left(-1, \frac{\pi}{2}\right)$ and perpendicular to $\sqrt{3} \sin \theta + 2 \cos \theta = \frac{4}{r}$ is:

If $PS$ is the median of the triangle with vertices $P(2,2)$,$Q(6,-1)$ and $R(7,3)$,then the equation of the line passing through $(1,-1)$ and parallel to $PS$ is

$A$ straight line passes through the point $A(3, 4)$ such that its intercept between the axes is bisected at $A$. Find its equation.

Vedclass Test Series

Exam Paper Generator

Online Exam Module