Question

find the equation line passing through the point of intersection of lines 2x+3y=13,5x-y-7=0and perpendicular to the line 3x-2y+7=0

Answer 1

first of all find the intersection points of lines.. supposed they come to be (a, b)
Now since line is perpendicular to 3x-2y+7=0
therefore slope product = -1
therefore, eqn of line is -3x-2y+c=0
Now put X=aand y=b in this eqn and find c..

Answer 2

Equation of line is 2x+3y=13

Given:

• 2x+3y=13 and 5x-y-7=0
• 3x-2y+7=0

To find:

• Find the equation of line passing through the point of intersection of lines 2x+3y=13,5x-y-7=0and perpendicular to the line 3x-2y+7=0.

Solution:

Formula to be used:

1. Equation of line passing through (x1,y1) having slope m $\bf (y - y_1) = m(x - x_1)$
2. Slope of two perpendicular lines are $\bf m_1m_2 = - 1$

Step 1:

Find intersection of 2x+3y=13 and 5x-y-7=0 to find (x1,y1)

Let

$2x + 3y = 13...eq1$

$5x - y = 7...eq2$

Multiply eq 2 by 3 and add eq1 and eq2

$2x + 3y = 13 \\ 15x - 3y = 21 \\ - - - - - - - - - \\ 17x = 34 \\ - - - - - - - - -$

or

$\bf x = 2$

put value of x in eq1

$2(2) + 3y = 13$

or

$3y = 13 - 4$

or

$y = \frac{9}{3}$

or

$\bf y = 3$

Thus,

line passing through (2,3).

Step 2:

Find slope of 3x-2y+7=0.

As

$- 2y = - 3x - 7$

or

$y = \frac{3}{2} x + \frac{7}{2}$

thus,

Slope is 3/2.

let slope of perpendicular line is m, so

$m \times \frac{3}{2} = - 1$

or

$\bf m = \frac{ - 2}{3}$

Step 3:

Find equation of line.

$(y - 3) = - \frac{2}{3} (x - 2)$

or

$3y - 9 = - 2x + 4$

or

$\bf 2x + 3y = 13$

Thus,

The equation line passing through the point of intersection of lines 2x+3y=13,5x-y-7=0and perpendicular to the line 3x-2y+7=0 is itself 2x+3y=13.

Learn more:

1) Given points A(1,5), B(-3, 7) and C(15,9).

(1) Find the equation of a line passing through the mid-point of AC and t...

https://brainly.in/question/12355170

2) Find the equation of the line passing through the point (-3.6, 2.1) and parallel to the line 4.9x + 5.4y = 3

https://brainly.in/question/45412085