Question

Find the equation of a straight line through the intersection of lines 5x-6y=2,3x+2y=10and perpendicular to the line 4x-7y+13=0

Answer 1

Answer:

The equation of the required line is 49x+28y-78=0

Step-by-step explanation:

Find the equation of a straight line through the intersection of lines 5x-6y=2,3x+2y=10and perpendicular to the line 4x-7y+13=0

$\text{The equation of the st.line pasing through the point of intesection of given two lines is }$

$5x-6y-2+\lambda(3x+2y-10)=0$

$(5+3\lambda)x+(-6+2\lambda)y+(-2-10\lambda)=0$........(1)

since (1) is perpendicular to 4x-7y+13=0,

$(\frac{-(5+3\lambda)}{(-6+2\lambda)})\times(\frac{4}{7})=-1$

$(\frac{(5+3\lambda)}{(-6+2\lambda)})\times(\frac{4}{7})=1$

$\frac{20+12\lambda}{-42+14\lambda}=1$

$20+12\lambda=-42+14\lambda$

$20+42=2\lambda$

$\implies\lambda=31$

$\text{The equation of the required line is } 98x+56y-312=0$

$\implies\boxed{49x+28y-78=0}$