Question

b) Find the equation of line which
passes through the point of
intersection of the lines x+2y-3=0
and 3x+4y-5=0 and which is
perpendicular to the line x-3y+5=0.​

Answer 1

First , we will find out the point of intersection of the lines x+2y-3=0 and 3x+4y-5=0.

Let point p(a,b) be point of intersection.

➝ x+2y-3=0

➝ a+2b-3=0

➝ a = 3-2b...... (1)

➝ 3x+4y-5=0

➝ 3a+4b-5=0

➝ a = (5-4b)/3.....(2)

From (1) and (2) :-

➝ 3-2b = (5-4b)/3

➝( 3-2b)3 = (5-4b)

➝ 9-6b = 5-4b

➝ 9-5 = 6b-4b

➝ 4 = 2b

➝ 2 = b

Now we will find the value of a.

=> a = 3-2b

=> a = 3-(2×2)

=> a = 3-4

=> a = -1

Point of intersection of the lines x+2y-3=0 and 3x+4y-5=0 = (-1,2)

Now , we will find Slope of line x-3y+5=0.

➝ x-3y+5=0

➝ -3y= -x -5

➝ y= (x + 5)/3

➝ y= (x/3) + ( 5/3)

Now we have transformed equation of line x-3y+5=0 as y= (x/3) + ( 5/3) which is of the form y = mx + c.

Therefore, Slope of line x-3y+5=0 = 1/3

Let. slope of line which passes through the point of intersection of the lines x+2y-3=0

and 3x+4y-5=0 and which is perpendicular to the line x-3y+5=0 be n.

➝ n × (1/3) = -1

[ product of slope of two perpendicular lines = -1 ]

➝ n = -1 × 3

➝ n = - 3

$\rm \: Equation \: of \: line \: passing \: through \: points \: (x_1 , y_1) \: and \: has \: slope \: \bold{m } :$

$\rm \longrightarrow y - y_1 =\bigg( x-x_1\bigg )m$

Now , we have to find equation of line which passes through point (-1,2) and has slope -3.

➝ y - 2 = (x+1)(-3)

➝ y - 2 = -3x-3

➝ y + 3x= -3+2

➝ y + 3x= - 1

➝ y + 3x + 1=0

Answer :

y + 3x + 1=0

Additional Information :

$\rm \large Equation \: of \: line \: passing \: through \: points \: (x_1 , y_1) \: and \: (x_2 , y_2) :$

$\rm\large \longrightarrow y - y_1 = x-x_1\bigg( \dfrac{y_2-y_1}{ x_2-x_1} \bigg )$

$\rm \large \rightarrow \: here \: \bigg( \dfrac{y_2-y_1}{ x_2-x_1} \bigg ) is \: slope \: of \: line$

Answer 2

${\large{\bold{\sf{\underline{Let's \; understand \; the \; question \; first}}}}}$

➨ This question says that we have to find the equation of line which passes through the point of intersection of the lines x+2y-3 = 0 and 3x+4y-5 = 0 and it is perpendicular to the line x-3y+5 = 0.

${\large{\bold{\sf{\underline{Given \; that}}}}}$

➨ Line passes the point of intersection of the lines x+2y-3 = 0 and 3x+4y-5 = 0

➨ It's perpendicular to the line x-3y+5 = 0.

${\large{\bold{\sf{\underline{To \; find}}}}}$

➨ Equation ( given question )

${\large{\bold{\sf{\underline{Solution}}}}}$

➨ Equation = y + 3x + 1 = 0

${\large{\bold{\sf{\underline{Assumptions}}}}}$

➨ Point p(r,m) is point of intersection

➨ a is the slope which passes through the point of intersection of the lines x+2y-3 = 0 and 3x+4y-5 = 0 and it is perpendicular to the line x-3y+5 = 0.

${\large{\bold{\sf{\underline{Full \; Solution}}}}}$

~ Firstly let us find the point of intersection of lines x+2y-3 = 0 and 3x+4y-5 = 0

➨ Point p(r,m) is point of intersection

➨ x + 2y - 3 = 0

➨ r + 2m - 3 = 0

➨ r = 3 - 2m Equation 1

➨ 3x + 4y - 5 = 0

➨ 3r + 4m - 5 = 0

➨ r = (5-4m) / 3 Equation 2

~ Now from equation 1 and 2

➨ 3 - 2m = (5-4m) / 3

➨ (3 - 2m)3 = (5-4m)

➨ 3(3 - 2m) = (5-4m)

➨ 9 - 6m = (5-4m)

➨ 9- 5 = 6m - 4m

➨ 4 = 2m

➨ 4/2 = m

➨ 2 = m

• Henceforth, the value of m is 2

~ Now let's find the value of r

➨ r = (3 - 2)m

➨ r = (3 - 2)2

➨ r = (3 - 2) × 2

➨ r = 3 - 4

➨ r = -1

• Henceforth, the value of r is -1

${\green{\frak{Henceforth, \: (-1,2) \: is \: intersecting \: point}}}$

~ Now let's see the slope

➨ x - 3y + 5 = 0

➨ -3y = -x - 5

➨ y = (x+5)/3

➨ y = (x/3) + (5/3)

➨ Henceforth, we get y = (x/3) + (5/3) at the place of x-3y+5=0. And it's a form of y = mx + c

➨ Therefore, slope of line is 1/3 now

~ Let's find the value of Assumption a

➨ a × 1/3 = -1

➨ a = -1 × 3

➨ a = -3

~ Now let's find the final result ( Equation )

➨ y - 2 = (x+1)(-3)

➨ y - 2 = -3x - 3

➨ y + 3x = -3 + 2

➨ y + 3x = -1

➨ y + 3x + 1 = 0

${\green{\frak{Henceforth, \: y + 3x + 1 = 0 \; is \; equation \; or \; final \; result}}}$