Question

find the equation of a straight line
Passing through the point of intersection
of lines, 5x-2y + 23 = 0 and 7x+6y-71=0 is perpendiculas to the
line Joining the points (5,1), (-2,2)​

Answer 1

Step-by-step explanation:

let the required line be y=mx+c

its perpendicular to 4x−2y=3

Slope of 4x−2y=3 ⇒ 4x−2y=3⇒ m

1= 2

4

=2

m

1

×m=−1

2×m=−1

m=−

2

1

x+c

∴ y=−

2

1

x+c

Passes through intersection of 5x−8y+23=0 and 7x+6y−71=0

3(5x−8y+23)=0⇒ 15x−24y+69=0

64(7x+6y−71)=0 ⇒ 28x+24y−284=0

---------------------------------------------------------------------------------------

43x=284−69

43x=215

x=

43

215

x=5

5(5)−8y+23=0

8y=48⇒y=6

meeting point P=(5,6)

P passes through y=mx+c

⇒ 6=

2

1

(5)+c

c=6+

2

5

=

2

17

∴ y=−

2

x

+

2

17

⇒ 2y+x−17=0

Equation of the line is 2y+x−17=0