The intercept form of the equation of the str | vedclass

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(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

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

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(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

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