The line passing through the points (1, 4) an | vedclass

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(C) The equation of the line passing through points $(x_1, y_1) = (1, 4)$ and $(x_2, y_2) = (-5, 1)$ is given by the two-point form: $\frac{y - y_1}{y

Vedclass · Accepted Answer

$(-1, 3)$

Vedclass · Answer

(C) The equation of the line passing through points $(x_1, y_1) = (1, 4)$ and $(x_2, y_2) = (-5, 1)$ is given by the two-point form: $\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$.Substituting the values: $\frac{y - 4}{1 - 4} = \frac{x - 1}{-5 - 1}$.$\frac{y - 4}{-3} = \frac{x - 1}{-6}$.$2(y - 4) = x - 1$ $\Rightarrow 2y - 8 = x - 1$ $\Rightarrow x - 2y + 7 = 0$ ...$(1)$.We are given the second line: $4x + 3y - 5 = 0$ ...$(2)$.From $(1)$,$x = 2y - 7$. Substituting this into $(2)$:$4(2y - 7) + 3y - 5 = 0$.$8y - 28 + 3y - 5 = 0$.$11y - 33 = 0 \Rightarrow y = 3$.Substituting $y = 3$ into $x = 2y - 7$: $x = 2(3) - 7 = 6 - 7 = -1$.Thus,the point of intersection is $(-1, 3)$.

The line passing through the points $(1, 4)$ and $(-5, 1)$ intersects the line $4x + 3y - 5 = 0$ at the point:

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