The equation of a line is the relationship between the x-coordinate and y-coordinate of any point on the line. The equation y = mx + b is the eqaution of straight line with slope m. Two line eqautions are perpendicular if they meet at right angles. Two lines are said to be perpendicular if the product of their slope is -1. If the lines y = m_{1}x + b_{1}, m_{1 }$\neq$ 0 and y = m_{2}x + b_{2}, m_{2} $\neq$ 0 are perpendicular, then m_{2} = $\frac{-1}{m_1}$.

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To find the perpendicular equation of any equation follow the procedure given below:

Interchange the coefficients of x and y in the equation of the given line, change the sign of either x or y and replace the constant term by a new constant k.

Solution:

Flip the coefficients of x and y in the equation of the given line and change the sign of x.

=> -2x + 5y = 10 (which is perpendicular equation to given equation)

Slope of perpendicular equation = $\frac{-coeff.\ of\ x}{coeff.\ of\ y}$ = $\frac{2}{5}$

=> Slope of a line perpendicular to the given line is $\frac{2}{5}$.

Let slope of given equation = $m_1$ and slope of perpendicular equation = $m_2$

=> $m_1$ = $\frac{-5}{2}$

and $m_2$ = $\frac{2}{5}$

Now $m_1$ * $m_2$ = $\frac{-5}{2}$ * $\frac{2}{5}$ = -1

=> $m_1$ * $m_2$ = -1

Product of both the slopes is -1. Hence lines are perpendicular.

If two lines are perpendicular to each other then the product of their slopes is -1 and conversely.

Proof :

Let AB and CD be two lines such that AB is perpendicular to CD having slopes m_{1} and m_{2} respectively.

Let the angle made by lines AB and CD with x-axis, in the positive direction be $\theta_1$ and $\theta_2$ respectively.

Then slope of AB = tan_{}$\theta_1$ = m_{1}

And slope of CD = tan_{}$\theta_2$ = m_{2}

From the above figure :

The inclination of CD exceeds AB by 90$^0$, that is Î¸_{2} = 90 + Î¸_{1}

Taking the tangent of both sides, we have

tan $\theta_2$_{} = tan ( 90 + $\theta_1$)

tan $\theta_2$ = - cot $\theta_1$ ($\because$ tan (90 + A) = - cot A)

tan $\theta_2$ = $\frac{-1}{ tan \theta_1}$

=> tan $\theta_2$ . tan $\theta_1$ = -1

or m_{1} . m_{2} = 1

Thus the product of the slopes = -1.

Two perpendicular lines cross each other at a right angle. If two lines have gradients $m_1$ and $m_2$ then the lines will be perpendicular if $m_1$ * $m_2$ = -1 and vice versa. To develop the equation of a line that is perpendicular to a given equation, simply flip the x and y coefficients and change one of the signs.

The standard form of the equation: ax + by = c

Flip 'a' and 'b' and change the sign of 'a'. The value of 'c' is not important, we can choose any value.

=> - bx + ay = c (or any value) is the perpendicular equation to the ax + by = c.

For example: Perpendicular equation to the given equation 18x + 5y = 11 is - 5x + 18y = 11

**Few examples based on perpendicular equations are discussed below:**

**Example:** Find the locus of the foot of the perpendicular from the origin to the line of which always passes through a fixed point (h, k).**Solution:** The equation of any line through (h, k) is

y - k = m (x - h) ----------------- (1)

The equation of the line perpendicular to it from the origin is

y = $\frac{-1}{m}$ x ------------------------ (2)

To get the locus of the point of intersection of (1) and (2), multiply both the equations

y (y - k) = - x(x - h)

or

x^{2} + y^{2} - hx - ky = 0,

Which is the required locus.

**Example 2**: Find the equation of a perpendicular line to 12x + 13y = 21, passing through the point (1, 5).

Solution:

Given equation is 12x + 13y = 21

Slope of equation = $\frac{-coeff.\ of\ x}{coeff.\ of\ y}$ = $\frac{-12}{13}$**The slopes of perpendicular lines are negative reciprocal to each other**.

=> Slope of perpendicular line is $\frac{13}{12}$

Now, equation of perpendicular line passing through (1, 5) is

y - 5 = $\frac{13}{12}$(x - 1)

12(y - 5) = 13(x - 1)

12y - 60 = 13x - 13

13x - 12y + 47 = 0

Which is the required equation.

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