Question

Find the equation of the straight line which passes through the point of intersection of the  straight lines x + y + 9 = 0 and 3x − 2y + 2 = 0 and is perpendicular to the straight line 4x+5y+1=0​

Answer 1

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

Given that,

A line passes through the point of intersection of the  straight lines x + y + 9 = 0 and 3x − 2y + 2 = 0.

We know,

↝ Any line which passes through the point of intersection of the  straight lines x + y + 9 = 0 and 3x − 2y + 2 = 0 is given by

$\rm :\longmapsto\:x + y + 9 + k(3x - 2y + 2) = 0$

↝ where, k is real number.

$\rm :\longmapsto\:x + y + 9 + 3kx - 2ky + 2k = 0$

$\rm :\longmapsto\:(1 + 3k)x + (1 - 2k)y + 9 + 2k = 0 - - - (1)$

↝ So, slope of line (1) is evaluated as

$\rm :\longmapsto\:Slope \: of \: line, \: m = - \: \dfrac{1 + 3k}{1 - 2k}$

$\rm :\longmapsto\:Slope \: of \: line, \: m = \: \dfrac{1 + 3k}{2k - 1}$

Now,

↝ It is given that, Line (1) is perpendicular to 4x + 5y + 1 = 0

↝ Let L represents the equation of line 4x + 5y + 1 = 0

So,

$\rm :\longmapsto\:Slope \: of \: line \: L = - \: \dfrac{4}{5}$

We know,

↝ Two lines having slope m and M are perpendicular iff Mm = - 1.

Thus,

$\rm :\longmapsto\: - \: \dfrac{4}{5} \times \dfrac{1 + 3k}{2k - 1} = - 1$

$\rm :\longmapsto\: \: \dfrac{4}{5} \times \dfrac{1 + 3k}{2k - 1} =1$

$\rm :\longmapsto\: \: \dfrac{4 + 12k}{10k - 5} =1$

$\rm :\longmapsto\:4 + 12k = 10k - 5$

$\rm :\longmapsto\:12k - 10k = - 5 - 4$

$\rm :\longmapsto\:2k = - 9$

$\rm :\longmapsto\:k = - \: \dfrac{9}{2}$

On substituting the value of k, in equation (1), we get

$\rm :\longmapsto\:\bigg(1 - \dfrac{27}{2} \bigg)x + (1 + 9)y + 9 - 9 = 0$

$\rm :\longmapsto\:\bigg( \dfrac{2 - 27}{2} \bigg)x + 10y = 0$

$\rm :\longmapsto\:\bigg( \dfrac{ -25 }{2} \bigg)x + 10y = 0$

$\rm :\longmapsto\: - 25x + 20y = 0$

$\rm :\longmapsto\: 5x - 4y = 0$

Additional Information :-

Different forms of equations of a straight line

1. Equations of horizontal and vertical lines

Equation of the line passes through (h, k) which is parallel to the x-axis is y = k.

and

Equation of line which is parallel to y-axis is x = h

2. Point-slope form

Consider a line whose slope is m and passes through the point ( a, b ), then equation of line is given by y - b = m(x - a)

3. Slope-intercept form

Consider a line whose slope is m which cuts the y-axis at a distance ‘a’ from the origin then equation of line is given by y = mx + a.

4. Intercept Form of Line

Consider a line having x– intercept a and y– intercept b, then the equation of line is x/a + y/b = 1.