For a square ABCD, the coordinates of two of its vertices are given as A (2, 5) and B (1, 3). The side length of the square is taken as ‘x’.
The coordinates for C and D will lie on which of the following lines?

2x - y - 4 =0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x - 2y + 5 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x - 2y - 1 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2x - y + 1 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2x + y + 1 =0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
2x - y - 4 =0

The line CD is parallel to AB and at a distance of $\sqrt{5}$ from the given line.
Slope of AB = 2.
So the equation of the line is of the form 2x – y – k = 0.
Distance between two parallel lines ax + by + c = 0 and ax + by + d =0 is $\frac{(c - d)}{\sqrt{a^{2} + b^{2}}}$ .
$∴$ k = 4, -6

Suggest Corrections

Similar questions

Q. Let

A B C D

be a square such that vertices

A, B, C, D

lie on circles

x^{2} + y^{2} - 2 x - 2 y = 0

x^{2} + y^{2} - 2 x - 2 y + 1 = 0

x^{2} + y^{2} + 2 x + 2 y + 1 = 0

and

x^{2} + y^{2} - 2 x + 2 y + 1 = 0

respectively with center of square being origin and sides are parallel to coordinate axes. The length of side of such square can be

If the line x+y-1 = 0 passes through the circumcentre and the point B of a triangle ABC, where $∠$ B = $90^{\circ}$ . Then the other two vertices (apart from B) of the triangle can lie on the line

Q. The equations of the tangents drawn from the point (0, 1) to the circle

x^{2} + y^{2} - 2 x + 4 y = 0

are

One bisector of the angle between the lines given by $a (x - 1)^{2} + 2 h (x - 1) y + b y^{2} = 0$ is 2x + y - 2 = 0.The other bisector is

In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

(a)

(b)

(c)

(d)

(e)

Join BYJU'S Learning Program

For a square ABCD, the coordinates of two of its vertices are given as A (2, 5) and B (1, 3). The side length of the square is taken as ‘x’. The coordinates for C and D will lie on which of the following lines?

The correct option is A 2x - y - 4 =0 The line CD is parallel to AB and at a distance of √5 from the given line. Slope of AB = 2. So the equation of the line is of the form 2x – y – k = 0. Distance between two parallel lines ax + by + c = 0 and ax + by + d =0 is (c−d)√a2+b2. ∴ k = 4, -6

For a square ABCD, the coordinates of two of its vertices are given as A (2, 5) and B (1, 3). The side length of the square is taken as ‘x’.
The coordinates for C and D will lie on which of the following lines?