Question

Find the foot of perpendicular drawn from (-2,-1) on the line 3x+ 2y-5=0​

Answer 1

Given :-

• A line 3x + 2y - 5 = 0 and a point ( - 2, - 1)

To Find :-

• Co-ordinates of foot of perpendicular drawn from (- 2, - 1) on the line 3x + 2y - 5 = 0.

Understanding the concept used :-

1. Slope of line :-

Let us consider a line ax + by + c = 0, then slope of line, m is given by

$\sf\: m \: = \: - \: \dfrac{coefficient \: of \: x}{coefficient \: of \: y}$

2. Condition for perpendicular lines :-

Let us consider two lines having slope m and M, then two lines are perpendicular iff m × M = - 1

3. Slope - point form of a line :-

Let us assume a line which passes through the point (a, b) and having slope 'm', then equation of line is given by (y - b) = m(x - a)

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

• Let assume that equation of line 3x + 2y - 5 = 0 ---(1) be denoted by L.

and

• Let the coordinate ( - 2, - 1 ) be denoted by P.

Now,

• Let PN be the perpendicular drop on Line L.

So, we have to find the coordinates of N.

Now,

• Equation of line L is 3x + 2y - 5 = 0,

So, Slope of line, L is given by

$\sf\: Slope \: of \: line, \: L = - \dfrac{coefficient \: of \: x}{coefficient \: of \: y} = - \dfrac{3}{2}$

Since,

• Line L is perpendicular to Line PN,

Therefore,

$\sf\:Slope \: of \: line, \: PN \: = \: \dfrac{2}{3}$

So,

• The equation of line PN, which passes through the point (-2, -1) having slope 2/3 is

$\sf\:y + 1 = \dfrac{2}{3} (x + 2)$

$\sf\: 3y + 3 = 2x + 4$

$\sf\:2x - 3y + 1 = 0 - - - (2)$

Now,

• To find the coordinates of foot of perpendicular, PN, we have to solve equation (1) and equation (2), using elimination method,

On multiply equation (1) by 2 and equation (2) by 3, we get

$\sf\: 6x + 4y - 10 \: = \: 0 - - - (3)$

and

$\sf \: 6x - 9y + 3 = 0 - - - (4)$

On Subtracting, equation (4) from equation (3), we get

$\sf \: 13y - 13 = 0$

$\sf \: 13y = 13$

$\bf \: \therefore \: y \: = \: 1 - - - (5)$

On substituting y = 1, in equation (2) we get

$\sf \: 2x - 3 \times 1 + 1 = 0$

$\sf \: 2x - 3 + 1 = 0$

$\sf \: 2x - 2 = 0$

$\sf \: 2x = 2$

$\bf \: \therefore \: x \: = \: 1 - - - (6)$

Hence,

• The coordinates of foot of perpendicular, N is (1, 1).

Additional Information

Different forms of equations of a straight line

1. Equations of horizontal and vertical lines

• Equation of the lines which are horizontal or parallel to the X-axis is y = a, where a is the y – coordinate of the points on the line.

• Similarly, equation of a straight line which is vertical or parallel to Y-axis is x = a, where a is the x-coordinate of the points on the line.

2. Point-slope form equation of line

• Consider a non-vertical line L whose slope is m, A(x,y) be an arbitrary point on the line and P(a, b) be the fixed point on the same line. Equation of line is given by y - b = m(x - a)

3. Slope-intercept form equation of line

• Consider a line whose slope is m which cuts the Y-axis at a distance ‘a’ from the origin. Then the distance a is called the y– intercept of the line. The point at which the line cuts y-axis will be (0,a). Then equation of line is given by y = mx + a.

4. Intercept Form of Line

• Consider a line L having x– intercept a and y– intercept b, then the line passes through  X– axis at (a,0) and Y– axis at (0,b). Equation of line is given by x/a + y/b = 1.

5. Normal form of Line

• Consider a perpendicular from the origin having length p to line L and it makes an angle β with the positive X-axis. Then, equation of line is given by x cosβ + y sinβ = p.