The equation of straight line equally inclined to the axes and equidistant from the point $(1, 2)$ and $(3, 4)$ is

x + y = 1

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y + 1 = 0

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y = x + 2

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y x + 1 = 0

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Solution

The correct option is C $y = x + 2$
Line is $y = x + c$
Distance from $(1, - 2)$ =distance from $(3, - 1)$
$\begin{matrix} \frac{\begin{matrix} | 1 + 2 - c | \end{matrix}}{\sqrt{2}} = \frac{\begin{matrix} | 4 - 3 - c | \end{matrix}}{\sqrt{2}} T h e r e f o r e, c = 2 ∴ y = x + 2 \end{matrix}$

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Similar questions

(1, - 2)

(3, 4)

(1, - 2)

(3, 4)

A (2, - 2)

B (6, 1)

C (- 3, 4)

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