Exercise 42 Page 162

To find our distance, we also need to know where the given line and the perpendicular line intersect. By setting up a system of equations, we can find the point of intersection.

{y = - \frac{a}{b} x y = \frac{b}{a} x + \frac{a y _{0} - b x _{0}}{a} (I) (II)

Since both equations have

y

isolated, it's convenient to use the Substitution Method. We will substitute the value of

y

from Equation (I) into Equation (II).

y = \frac{b}{a} x + \frac{a y _{0} - b x _{0}}{a}

Substitute

$y = - \frac{a}{b} x$

- \frac{a}{b} x = \frac{b}{a} x + \frac{a y _{0} - b x _{0}}{a}

SubEqn

$LHS - \frac{b}{a} x = RHS - \frac{b}{a} x$

- \frac{a}{b} x - \frac{b}{a} x = \frac{a y _{0} - b x _{0}}{a}

FactorOut

Factor out $x$

x (- \frac{a}{b} - \frac{b}{a}) = \frac{a y _{0} - b x _{0}}{a}

DivEqn

$LHS / (- \frac{a}{b} - \frac{b}{a}) = RHS / (- \frac{a}{b} - \frac{b}{a})$

x = \frac{( \frac{a y _{0} - b x _{0}}{a} )}{( - \frac{a}{b} - \frac{b}{a} )}

From here, we subtract the fractions in the denominator before multiplying the numerator and the denominator by

a b .

x = \frac{( \frac{a y _{0} - b x _{0}}{a} ) \cdot a b}{( - \frac{a}{b} - \frac{b}{a} ) \cdot a b} \Leftrightarrow x = \frac{b ( a y _{0} - b x _{0} )}{- a ^{2} - b ^{2}}

Now, we simplify a little more by multiplying the numerator and denominator by

- 1 .

x = \frac{b ( a y _{0} - b x _{0} ) \cdot ( - 1 )}{( - a ^{2} - b ^{2} ) \cdot ( - 1 )} \Leftrightarrow x = \frac{b ( b x _{0} - a y _{0} )}{a ^{2} + b ^{2}}

Having solved the second equation for

x,

we can substitute this value into the first equation to find the value of

y .

y = - \frac{a}{b} x

Substitute

$x = \frac{b ( b x _{0} - a y _{0} )}{a ^{2} + b ^{2}}$

y = - \frac{a}{b} \cdot \frac{b ( b x _{0} - a y _{0} )}{a ^{2} + b ^{2}}

▼

Simplify right-hand side

MultFrac

Multiply fractions

y = - \frac{a b ( b x _{0} - a y _{0} )}{b ( a ^{2} + b ^{2} )}

ReduceFrac

$\frac{a}{b} = \frac{a / b}{b / b}$

y = - \frac{a ( b x _{0} - a y _{0} )}{a ^{2} + b ^{2}}

MoveNegFracToNum

Put minus sign in numerator

y = \frac{- a ( b x _{0} - a y _{0} )}{a ^{2} + b ^{2}}

DistrNegSignSwap

$- (b - a) = a - b$

y = \frac{a ( a y _{0} - b x _{0} )}{a ^{2} + b ^{2}}

Finding the Distance

When we know the coordinates of the two points making up the segment, we can use the Distance Formula to calculate it.

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

SubstitutePoints

Substitute $(\frac{b ( b x _{0} - a y _{0} )}{a ^{2} + b ^{2}}, \frac{a ( a y _{0} - b x _{0} )}{a ^{2} + b ^{2}})$ & $(x_{0}, y_{0})$

d = (\frac{b ( b x _{0} - a y _{0} )}{a ^{2} + b ^{2}} - x_{0})^{2} + (\frac{a ( a y _{0} - b x _{0} )}{a ^{2} + b ^{2}} - y_{0})^{2}

▼

Simplify right-hand side

NumberToFrac

$a = \frac{( a ^{2} + b ^{2} ) \cdot a}{( a ^{2} + b ^{2} )}$

d = (\frac{b ( b x _{0} - a y _{0} )}{a ^{2} + b ^{2}} - \frac{x _{0} ( a ^{2} + b ^{2} )}{a ^{2} + b ^{2}})^{2} + (\frac{a ( a y _{0} - b x _{0} )}{a ^{2} + b ^{2}} - \frac{y _{0} ( a ^{2} + b ^{2} )}{a ^{2} + b ^{2}})^{2}

Distr

Distribute $coefficients$

d = (\frac{b ^{2} x _{0} - a b y _{0}}{a ^{2} + b ^{2}} - \frac{a ^{2} x _{0} + b ^{2} x _{0}}{a ^{2} + b ^{2}})^{2} + (\frac{a ^{2} y _{0} - a b x _{0}}{a ^{2} + b ^{2}} - \frac{a ^{2} y _{0} + b ^{2} y _{0}}{a ^{2} + b ^{2}})^{2}

SubFrac

Subtract fractions

d = (\frac{- a b y _{0} - a ^{2} x _{0}}{a ^{2} + b ^{2}})^{2} + (\frac{- a b x _{0} - b ^{2} y _{0}}{a ^{2} + b ^{2}})^{2}

To simplify this further, there are two things we need to do. First, remember that the square of a negative is a positive number. Then, we can use the properties of perfect square trinomials.