A Point Equidistant From The Lines 4x + 3y + 10 = 0, 5x – 12y + 26 = 0 And 7x + 24y – 50 = 0 Is

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Question

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

Solution

Correct option is

(0, 0)

42 +32 = 52, 52 + 122 = 132, 72 + 242 = 252

Hence the distance of (0, 0) form the three sides are 

Testing

SIMILAR QUESTIONS

Q1

 

Given four lines with equations x + 2y – 3 = 0, 3x + 4y – 7 = 0,

2x + 3y – 4 = 0, 4x + 5y – 6 = 0, then

Q2

Area of the parallelogram formed by the lines y = mx, y = mx + 1, y = nxand y = nx + 1 equals.

Q3

The area bounded by the curves 

Q4

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

Q5

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

Q6

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

Q7

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

Q8

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 

Q9

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

Q10

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