Question

The Line segment joining P(5,-2) and Q(9,6) is divided in the ratio 3:1 by a point A on it. Find the equation of a line through the point A perpendicular to the line x-3y+4=0.​

Answer 1

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

Given that,

The line segment joining P(5,-2) and Q(9,6) is divided in the ratio 3:1 by a point A on it.

Let assume that coordinates of A be (a, b).

We know,

Section Formula

Let us consider a line segment joining the points A and B and let C (x, y) be any point which divides AB internally in the ratio m : n, then coordinates of C is given by

$\boxed{\bf \:( x, y) = \bigg(\dfrac{mx_2 + nx_1}{m + n} , \dfrac{my_2 + ny_1}{m + n}\bigg)}$

So, here

$\rm :\longmapsto\:x_1 = 5,y_1 = - 2,x_2=9 ,y_2=6 ,m = 3,n =1$

So, on substituting the values, we get

$\rm :\longmapsto \:( a, b) = \bigg(\dfrac{3 \times 9 + 1 \times 5}{3 + 1} , \dfrac{3 \times 6 - 2 \times 1}{3 + 1} \bigg)$

$\rm :\longmapsto \:( a, b) = \bigg(\dfrac{27 + 5}{4} , \dfrac{18 - 2}{4} \bigg)$

$\rm :\longmapsto \:( a, b) = \bigg(\dfrac{32}{4} , \dfrac{16}{4} \bigg)$

$\rm :\longmapsto \:( a, b) = (8 ,4)$

So, Coordinates of A is ( 8, 4 ).

Let assume that

'l' represents the equation of line is x - 3y + 4 = 0.

We have to find the equation of line 'L' which passes through A ( 8, 4 ) and perpendicular to line 'l'.

Now,

Given that,

Equation of line, l is x - 3y + 4 = 0

So,

$\rm :\longmapsto\:Slope \: of \: l \: = - \: \dfrac{coefficient \: of \: x}{coefficient \: of \: y}$

$\rm :\longmapsto\:Slope \: of \: l \: = - \: \dfrac{1}{( - 3)}$

$\rm :\longmapsto\:Slope \: of \: l \: = \: \dfrac{1}{3}$

Since, L is perpendicular to L.

We know that, two lines having slope m and M are perpendicular iff Mm = - 1

So,

$\rm :\longmapsto\:Slope \: of \: L, \: m = - 3$

So,

Equation of line L, passes through the point A ( 8, 4 ) having slope m = - 3 is given by

$\rm :\longmapsto\:y - 4 = - 3(x - 8)$

$\rm :\longmapsto\:y - 4 = - 3x + 24$

$\bf :\longmapsto\:3x + y - 28 = 0$

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.