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Question

Mathematics Question on general equation of a line

If each line of a pair of lines passing through origin is at a perpendicular distance of 44 units from the point (3,4)(3, 4), then the equation of the pair of lines is

A

7x2+24xy=07x^2 + 24xy = 0

B

7y2+24xy=07y^2 + 24xy = 0

C

7y224xy=07y^2 - 24xy = 0

D

7x224xy=07x^2 - 24xy = 0

Answer

7y2+24xy=07y^2 + 24xy = 0

Explanation

Solution

Let equation of line passes through origin having slope mm is ymx=0y-m x=0, according to given information
43m1+m2=4\frac{|4-3 m|}{\sqrt{1+m^{2}}}=4
16+9m224m=16+16m2\Rightarrow 16+9 m^{2}-24 m=16+16 m^{2}
7m2+24m=0\Rightarrow 7 m^{2}+24 m=0
m=0\Rightarrow m=0 or m=247m=-\frac{24}{7}
so combined equation of required lines
y(y+247x)=0y\left(y+\frac{24}{7} x\right)=0
7y2+24xy=0\Rightarrow 7 y^{2}+24 x y=0