The distance between the point (1, 2) and a point on the line x+y = 6 is 3 units. The coordinates of the point is

(3,3)

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(4,2)

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(-1,7)

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(1,3)

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Solution

The correct option is B
(4,2)

Any point on the given line can be taken as (x, 6-x) Given that the distance between the points is 3 units

Distance formula = $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

Using both the relations we get the coordinates as (4, 2) and (1, 5).

Suggest Corrections

Similar questions

The distance between the point (1, 2) and a point on the line x+y = 6 is 3 units. The coordinates of the point is

\frac{x + 1}{2} = \frac{y + 2}{3} = \frac{z - 3}{6}

3

(1, - 2, 3)

The coordinates of a point at unit distance from the lines 3x - 4y + 1 = 0 and 3x + 6y + 1 = 0 are

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z}{\sqrt{3}}

1

(- 1, - 1, - \sqrt{3})

A (2, - 3, - 1)

B (8, - 1, 2)

14

A

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