Question

Find the equation of a straight line passing through the point (-3,1) and perpendicular to the line 5x-2y+7=0​

Answer 1

Answer:

Let the equation of line which is perpendicular to

5x−2y=7 is 2x+5y=λ ...... (i)

The given lines are 2x+3y=1 ...... (ii) and 3x+4y=6 ..... (iii)

Solving (ii) and (iii), we get

x=14,y=−9

∴ The point of intersection of given lines is (14,−9)

Since, the Eq. (i) is passing through the point (14,−9)

∴2(14)+5(−9)=λ⇒λ=−17

∴ Eq. (i) becomes

2x+5y+17=0

Step-by-step explanation:

Answer 2

Given line is 5x - 2y +7 = 0

=> 2y = 5x + 7

=> y = (5/2)x + 7/2

ie, slope m = 5/2

Since 2 parallel lines have the same slope

So, slope of the unknown line = 5/2

Now, using slope formula:

y- y1 = m( x - x1)

Since unknown line passes through ( -3, 4)

=> x1 = -3 & y1 = 4

=> eq: y - 4 = (5/2) ( x- -3 )

=> y-4 = 5x/2 + 15/2

=> 2y - 8 = 5x + 15

=> 5x - 2y = -23

So, required equation is 5x -2y = -23