Question

Find the equation of the line through the intersection of of x - 3 y = 1 and 2 X + 3 y - 23 = 0 and perpendicular to the line 5 x - 3 y - 1 = 0

Answer 1

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

The required equation of line which passes through the point of intersection of the lines x - 3y = 1 and 2x + 3y - 23 = 0 is

$\rm :\longmapsto\:x - 3y - 1 + k(2x + 3y - 23) = 0$

$\rm :\longmapsto\:x - 3y - 1 + 2kx + 3ky - 23k= 0$

$\rm :\longmapsto\:(1 + 2k)x + (3k - 3)y - 1 - 23k= 0 - - - (1)$

Now, it is given that line 1 is perpendicular to the line 5x - 3y - 1 = 0

So,

$\boxed{\tt{ Slope \: of \: (1) = - \frac{1 + 2k}{3k - 3} = \frac{1 + 2k}{3 - 3k} \: }}$

$\boxed{\tt{ Slope \: of \: 5x - 3y - 1 = 0 \: is \: - \dfrac{5}{ - 3} = \dfrac{5}{3} }}$

We know,

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

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

$\rm \implies\:\dfrac{5 + 10k}{9 - 9k} = - 1$

$\rm \implies\:5 + 10k = 9k - 9$

$\rm \implies\:10k - 9k = - 9 - 5$

$\rm \implies\:k = - 14$

On substituting k = - 14 in equation (1), we get

$\rm :\longmapsto\:(1 - 28)x + ( - 42 - 3)y - 1 + 322= 0$

$\rm :\longmapsto \: - 27x- 45y + 321= 0$

$\rm :\longmapsto \: - 3(9x + 15y - 107)= 0$

$\rm :\longmapsto \: 9x + 15y - 107= 0$

is the required equation of line which passes through the point of intersection of the lines x - 3y = 1 and 2x + 3y - 23 = 0 and perpendicular to the line 5x - 3y - 1 = 0

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Different forms of equations of a straight line

1. Equations of horizontal and vertical lines

Equation of line parallel to x - axis passes through the point (a, b) is x = a.

Equation of line parallel to x - axis passes through the point (a, b) is x = a.

2. Point-slope form equation of line

Equation of line passing through the point (a, b) having slope m is y - b = m(x - a)

3. Slope-intercept form equation of line

Equation of line which makes an intercept of c units on y axis and having slope m is y = mx + c.

4. Intercept Form of Line

Equation of line which makes an intercept of a and b units on x - axis and y - axis respectively is x/a + y/b = 1.

5. Normal form of Line

Equation of line which is at a distance of p units from the origin and perpendicular makes an angle β with the positive X-axis is x cosβ + y sinβ = p.