Question

The equation of the straight line which passes through the point (1, –2) and cuts off equal intercepts from the axes will be

Solution

Correct option is

x + y + 1 = 0

Line cutting equal intercepts on axes is of the from x + y = c and

         x – y = c.

it passes through (1, –2)

      x + y + 1 = 0 or  x – y – 3 = 0,

SIMILAR QUESTIONS

Q1

A line making an angle  with the + ive direction of x-axis passes throughP(5, 6) to meet the line x = 6 at Q and y = 9 at R the QR is

Q2

A straight line through the origin O meets the parallel lines 4x + 2y = 9 and 2x + y + 6 = 0 at points P and Q respectively. Then the point Odivides the segment PQ in the ratio

Q3

Orthocenter of triangle whose vertices are (0, 0), (3, 4), (4, 0) is

Q4

The three lines 4x – 7y + 10 = 0, x + y = 5 and 7x + 4y = 15 from the sides of a triangle. The line (1, 2) is its

Q5

The equation to the side of a triangle are x – 3y = 0, 4x + 3= 5 and 3x + y = 0. The line 3x – 4y = 0 passes through 

Q6

One vertex of the equilateral triangle with centroid at the origin and one side as x + y – 2 = 0 is

Q7

A point equidistant from the lines 4x + 3y + 10 = 0, 5x – 12y + 26 = 0 and 7x + 24y – 50 = 0 is

Q8

Let P = (–1, 0), Q = (0, 0) and  be three points. Then the equation of the bisector of the angle PQR is

Q9

Area of Δ formed by line x + y = 3 and  bisectors of pair of straight lines x2 – y2 +2y = 1is

Q10

The three lines 3x + 4y + 6 = 0,  and 4x + 7y + 8 = 0 are