Question

Equation of straight line passing through the point (4, 5) and equally inclined to the lines 3x=4y+7 and 5y=12x+6 is?

Answer 1

let the slope of the line be 'm' and equate both the tan∅ i.e. [( m1-m2)/1+m1m2] (angle betⁿ two lines.) where m1 & m2 are the slopes of the given lines ...from here u get 'm'. let the eqⁿ of line be. y=mx+c put the given y & x and 'm' u will get 'c'. hence u will got the ans.
if satisfied. mark as brainliest...

Answer 2

Answer:

7x+9y=73

Step-by-step explanation:

let the slope of line be m1

3x=4y+7

y=(34)x+74so m=34 eqn(1)

y=125x+65 so m=125 eqn(2)

θ=tan−1∣∣∣m1−m21+m1m2∣∣∣

=tan−1(m1−341+34m1)=tan−1(m1−1251+125m1)

=tan−1(4m1−34+3m1)=tan−1(5m1−125+12m1)

=4m1−31+3m1=5m1−125+12m1

=(4m1−3)(5+12m1)=(5m1−12)(4+3m1)

=20m1−15+48m21−36m1=20m1−48+15m21−36m1

=33m21=−33

m21=−1

this cant be possible so changing the signs of equation

20m1−15+48m21−36m1=−20m1+48−15m1+36m1

63m21−32m1−63=0

m1=32±130126

m1=162126,−98126

m1=5443,−79

putting it back in the equations

y−5=5443(x−4)

43y−215=54x−216

54x−43y=1 answer

also, y−5=−79(x−4)

5y−45=−7x+28

7x+9y=73 answer