Q7 of 7 Page 211

The distance of the point (3 5) from the line 2x + 3y–14 = 0 measured parallel to the line x–2y = 1 is:

a) 7


b) 13


c) 5


d) 3

The given line is 2x + 3y = 14 = L1


Line parallel to x –2y = 1 which passes through (3,5) is 3–2(5) = k


K = –7


So, equation of line becomes x –2y = –7 = L2


The intersection point of L1 and L2


2x + 3y = 14


x –2y = –7


Multiplying L2 by 2 and the n L1 – L2, we get,


x = 1 and y = 4


So, distance between (1,4) and (3,5)


Using distance formula



So, distance of the point (3, 5) from the line 2x + 3y–14 = 0 measured parallel to the line x–2y = 1 is 5 units


More from this chapter

All 7 →
3

Express y in terms of x given that 2x–5y = 7. Check whether the point (–3 –2) is on the given line.

 4

Each side of a square is of length 6 units and the centre of the square is (–1 2). One of its diagonals is parallel to x + y = 0. Find the coordinates of the vertices of the square.

 5

The area of triangle formed by the line with the coordinate axes is-

a) ab


b)2ab


c)1/2ab


d)2/2ab

 6

If AB = x + 3, BC = 2x, and AC = 4x–5 then for what value of x, B lies on AC?