Question

Find the equation of the line through the point of intersection of 2x-3y+1=0 and x+y-2=0 which is parallel to the x axis

Answer 1

Step-by-step explanation:

Step 1 :

Let the equation of the line passing through the point of intersection be

(x−y+1)+λ(2x−3y+5)=0

( i.e) x(1+2λ)−y(1+3λ)+1+5λ=0

Step 2 :

Its distance from the point (3,2) is given as 75

∴∣∣∣3(1+2λ)−2(1+3λ)+1+5λ(1+2λ)2+(1+3λ)2−−−−−−−−−−−−−−−√∣∣=75

⇒∣∣∣3+6λ−2−6λ+1+5λ1+4λ2+4λ+1+6λ+9λ2−−−−−−−−−−−−−−−−−−−√∣∣=75

⇒∣∣∣2+5λ2+10λ+13λ2−−−−−−−−−−−√∣∣=75

Step 3 :

Squaring on both sides we get,

(2+5λ)213λ2+10λ+2=4925

4+20λ+25λ213λ2+10λ+2=4925

25(4+20λ+25λ2)=49(13λ2+10λ+2)

100+500λ+625λ2=637λ2+490λ+98

⇒12λ2−10λ−2=0

⇒6λ2−5λ−1=0

On factorizing we get,

(6λ+1)(λ−1)=0

Hence λ=−16 or λ=1

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Step 4 :

The equation of the required line is when λ=−16

x(1+2(−16))−y(1+3(−16))+1+5(−16)=0

(i.e) x(23)−y(12)+16=0 (i.e) 4x−3y+1=0

Case (ii)

when λ=1

x(1+2(1))−y(1+3(1))+1+5(1)=0

x(3)−y(4)+6=0

3x−4y+6=0

Hence the equations are

4x−3y+1=0 and 3x−4y+6=0