Question

12. Find the equation of the straight line which passes through the point of intersection of the straight lines 5x-6y=1 and 3x+2y+5=0 and is perpendicular to the straight line 3x-5y+11 = 0.

Answer 1

$\textbf{\underline{Answer :}}$

The two given lines are

5x - 6y = 1 ...(i)

and

3x + 2y + 5 = 0

⇒ 3x + 2y = - 5 ...(ii)

Now, (i) × 3 and (ii) × 5 ⇒

15x - 18y = 3

15x + 10y = - 25

On subtraction, we get

- 18y - 10y = 3 + 25

⇒ - 28y = 28

⇒ y = - 1

Now, putting y = - 1 in (i), we get

5x - 6 (- 1) = 1

⇒ 5x + 6 = 1

⇒ 5x = 1 - 6

⇒ 5x = - 5

⇒ x = - 1

So, the intersection of the lines (i) and (ii)
is (- 1, - 1)

Any perpendicular line to the line 3x - 5y + 11 = 0 is

5x + 3y = k ...(iii)

The point (- 1, - 1) lies on the line (iii). So,

5 (- 1) + 3 (- 1) = k

⇒ - 5 - 3 = k

⇒ k = - 8

Hence, from (iii), we get the required line as

5x + 3y = - 8

i.e., 5x + 3y + 8 = 0

#$\textbf{MarkAsBrainliest}$

Answer 2

Solution :

i ) Given two straight lines ,

5x - 6y = 1 ---( 1 )

3x + 2y = -5 ----( 2 )

Let the intersecting point P(x,y ).

Finding intersecting point :

multiply ( 2 ) with 3 , and add

equation ( 1 ) , we get

14x = -14

=> x = -14/14 = -1

substitute x = -1 in equation ( 1 ) , we get

5( -1 ) - 6y = 1

=> -5 - 6y = 1

=> -6y = 6

=> y = -6/6 = -1

P( x , y ) = ( -1 , -1 )

ii ) Slope of a line 3x -5y +11 = 0

m1 = -a/b = -3/(-5) = 3/5

slope of a line perpendicular to

above line ( m2 ) = - 1/m1

m2 = -5/3

iii ) Required equation , whose

slope ( m2) = -5/3 , and passing

through the point P( -1,-1 ) = ( x1 , y1 )

is

y - y1 = m2( x - x1 )

=> y +1 = ( -5/3 ) ( x + 1 )

=> 3(y+1) = -5(x+1)

=> 5(x+1)+3(y+1) = 0

=> 5x + 5 + 3y + 3 = 0

=> 5x + 3y + 8 = 0

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