Question

find the equation of line passing through the point of intersection of the line x-y-1=0 and 2x-3y+1=0 and parallel to the line 3x+4y-13=0



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Answer 1

Question :--- find the equation of line passing through the point of intersection of the line x-y-1=0 and 2x-3y+1=0 and parallel to the line 3x+4y-13=0 ..

Formula used :---

→ Slope of line in the form of y = mx + c is given by m.

→ when lines are parallel than m1*m2 = (-1)

→ Equation of line is given by = slope = (y-y1)/(x-x1)

_____________________________

Solution :---

→ x - y - 1 = 0

→ x - y = 1 --------- Equation (1)

and,

→ 2x-3y + 1 = 0

→ 2x - 3y = (-1) -------- Equation (2)

Multiplying Equation (1) by 2 we get,

→ 2x - 2y = 2

Now, subtracting Equation(2) From Equation (1) we get,

→ (2x-2y) - (2x-3y) = 2 - (-1)

→ y = 3

Putting in Equation (1) now we get,

→ x - 3 = 1

→ x = 4..

Therefore, point of intersection = (4,3)

Now, It is parallel to 3x+4y = 13

→ 4y = 13 - 3x

→ y = -3x/4 + 13/4

→ y = mx + c

Therfore (m) slope = -3/4 ....

So, Equation of line will be :---

→ (-3/4) = (y - 3)/(x - 4)

Cross - Multiply we get,

→ 4y - 12= (-3)x + 12

→ 3x + 4y = 24

→ 3x + 4y - 24 = 0

Hence, the Equation of line will be 3x + 4y - 24 = 0.

Answer 2

Answer:

${\boxed{\bold{Equation : 3x + 4y - 24 = 0 }}}$

Step-by-step explanation:

Given equations,

• x - y - 1 = 0
• 2x - 3y + 1 = 0
• Parallel to line : 3x + 4y - 13 = 0
• Equation of line = ?

At first,

$\rm x - y - 1 = 0 \\\rightarrow\rm x - y = 1 \\\rightarrow\rm x = y + 1 ........(eq.1)$

Secondly,

$\rm 2x - 3y + 1 = 0 \\\rightarrow\rm 2x - 3y = - 1 \\\rightarrow\rm 2x = 3y - 1 \\\tt * Putting \: values\: of \:(eq.1):- \\\rightarrow\rm 2(y + 1) = 3y - 1 \\\rightarrow\rm 2y + 2 = 3y - 1 \\\rightarrow\rm 3y - 2y = 2 + 1 \\\rightarrow{\boxed{\rm{y = 3 }}}$

$\tt * Putting \:values\: of\:y \: in\:(eq.1)$

$\rightarrow\rm x = 3 + 1 \\\rightarrow{\boxed{\rm{x = 4}}}$

$\therefore\tt Point\:of\:intersection(x,y) = (4,3)$

$\therefore\sf x - 4 = 0 \: \: and\: \: y - 3 = 0$

$\tt Now, it \:is\:parallel\:to\: the\: line : 3x + 4y - 13 = 0$

$\rightarrow\rm 4y = -3x - 13 \\\rightarrow\rm y =\frac{-3x-13}{4} \\\rightarrow\rm y = \frac{-3x}{4} + \frac{-13}{4} \\\rightarrow\rm y =\frac{-3}{4}x + \frac{-13}{4} \\\rightarrow\rm y = mx + c \: \: [Where,m = \frac{-3}{4}]$

$\therefore\tt Value\:of\:slope(m) =\frac{-3}{4}$

$\rm So, \frac{-3}{4} = \frac {y -3}{x-4} \\\rightarrow\rm 4(y- 3)= -3(x - 4) \\\rightarrow\rm 4y - 12 = - 3x + 12 \\\rightarrow\rm 4y + 3x = 12 + 12 \\\rightarrow\rm 3x + 4y = 24 \\\rightarrow{\boxed{\rm {3x + 4y - 24 = 0}}}$

$\therefore\bold{\underline{Equation\:of\:line=3x + 4y - 24 = 0 }}$