If each of the points (a, 4) and (-2, b) lies | vedclass

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(C) The equation of the line passing through $(2, -1)$ and $(5, -3)$ is given by the two-point form: $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.

Vedclass · Accepted Answer

$2x + 6y + 1 = 0$

Vedclass · Answer

(C) The equation of the line passing through $(2, -1)$ and $(5, -3)$ is given by the two-point form: $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.Substituting the points $(2, -1)$ and $(5, -3)$:$y - (-1) = \frac{-3 - (-1)}{5 - 2}(x - 2)$$y + 1 = \frac{-2}{3}(x - 2)$$3(y + 1) = -2(x - 2)$$3y + 3 = -2x + 4$$2x + 3y = 1$ (Equation $L$)Since $(a, 4)$ lies on $L$:$2(a) + 3(4) = 1$ $\Rightarrow 2a + 12 = 1$ $\Rightarrow 2a = -11$ $\Rightarrow a = -\frac{11}{2}$.Since $(-2, b)$ lies on $L$:$2(-2) + 3(b) = 1$ $\Rightarrow -4 + 3b = 1$ $\Rightarrow 3b = 5$ $\Rightarrow b = \frac{5}{3}$.We need to check which line contains the point $(a, b) = (-\frac{11}{2}, \frac{5}{3})$.Testing option $C$: $2x + 6y + 1 = 0$$2(-\frac{11}{2}) + 6(\frac{5}{3}) + 1 = -11 + 10 + 1 = 0$.Thus,the point $(a, b)$ lies on the line $2x + 6y + 1 = 0$.

If each of the points $(a, 4)$ and $(-2, b)$ lies on the line joining the points $(2, -1)$ and $(5, -3)$,then the point $(a, b)$ lies on the line:

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