Question

the equation of line passing through intersection of the line x-2y+5=0 and 3x +2y+7=0 and perpendicular to x-y=0​

Answer 1

Answer:

x+y+2=0

Step-by-step explanation:

X-2Y+5=0-------------------->eq(1)

3X+2Y+7=0------------------->eq(2)

perpendicular line

:.X-Y=0.

slope

ax+by+c=0

M=-a/b

line (X-Y=0)

M=-1/-1=1

:.M=1------->1st slope..

perpendicular lines implies=>product of slopes

M1M2=-1

=>1*M2=-1

=>M2=-1

X-2Y+5=0-------------------->eq(1)

3X+2Y+7=0------------------->eq(2)

cancel those 2y's

then we will get this eqn

:.4x+12=0

4x=-12

X=-3

x value substitute in eq(1)

-3x-2y+5

-2y+2=0

-2y =-2

y=1

point intersection p(-3,1)

M=-1 by seen line eqn X-Y=0

point slope

y-y1=m(x-x1)

substitute X1,y1 in point intersection (-3,1)

y-1=-1(x-(-3))

y-1=-1(x+3)

y-1=-x-3

x+y-1+3=0

by left hand side 3

x+y+2=0.

line eqn