Q. 55.0( 2 Votes )
Show that the origin is equidistant from the lines 4x + 3y + 10 = 0; 5x – 12y + 26 = 0 and 7x + 24y = 50.
Answer :
Given: The lines 4x + 3y + 10 = 0; 5x – 12y + 26 = 0 and 7x + 24y = 50.
To prove:
The origin is equidistant from the lines 4x + 3y + 10 = 0; 5x – 12y + 26 = 0 and 7x + 24y = 50.
Explanation:
Let us write down the normal forms of the given lines.
First line: 4x + 3y + 10 = 0
⇒ - 4x - 3y = 10
Dividing both sides by 
∴ p = 2
Second line: 5x − 12y + 26 = 0
⇒ - 5x + 12y = 26
Dividing both sides by 
∴ p = 2
Third line: 7x + 24y = 50
Dividing both sides by 
∴ p = 2
Hence, the origin is equidistant from the given lines.
Rate this question :
Various Forms of Equations of line45 mins
Parametric Equations of Straight line48 mins
Slope, inclination and angle between two lines48 mins
Interactive Quiz on Equations of line23 mins
Straight line | Analyse your learning through quiz56 mins
General Equation of a line43 mins
Motion in a Straight Line - 0665 mins
Motion in a Straight Line - 0556 mins
Motion in a Straight Line - 0372 mins
Understand The Interesting Concept Of Relative velocity in 1- Dimension44 mins
Find the values of θ and p, if the equation x cos θ + y sin θ = p is the normal form of the line 
.
The vertices of a triangle are (6, 0), (0, 6) and (6, 6). The distance between its circumcentre and centroid is
RD Sharma - MathematicsFor specifying a straight line, how many geometrical parameters should be known?
Mathematics - ExemplarIf the line 
passes through the points (2, –3) and (4, –5), then (a, b) is
