Chapter 13.4, Problem 31WE

To determine

To prove: The given expression by using algebra.

Explanation of Solution

Given information: The given expression is $\left|\left(ka,kb\right)\right|=\left|k\right|\cdot \left|\left(a,b\right)\right|$ .

Calculation:

Absolute value is the magnitude of any given real number irrespective of other values and signs.

Formula to determine distance between two points is:

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Let, the one point through which line crosses have coordinates $\left(ka,kb\right)$ and the other point be at origin, that is, $\left(0,0\right)$ .

To determine the value of left hand side expression of the given expression, substitute $ka$ for $x_2$ , $kb$ for $y_2$ , $0$ for $x_1$ and $0$ for $y_1$ in the above distance formula.

  $\begin{array}{c}L.H.S.=\left|\left(ka,kb\right)\right|=\sqrt{{\left(ka-0\right)}^2+{\left(kb-0\right)}^2}\\ =\sqrt{k^2{a}^2+k^2{b}^2}\\ =\sqrt{k^2\left(a^2+b^2\right)}\\ =k\sqrt{\left(a^2+b^2\right)}\end{array}$

To determine the value of right hand side expression of the given expression, substitute $a$ for $x_2$ , $b$ for $y_2$ , $0$ for $x_1$ and $0$ for $y_1$ in the above distance formula.

  $\begin{array}{c}\text{R}\text{.H}\text{.S}\text{.}=\left|k\right|\cdot \left|\left(a,b\right)\right|=k\sqrt{{\left(a-0\right)}^2+{\left(b-0\right)}^2}\\ =k\sqrt{\left(a^2+b^2\right)}\end{array}$

The left hand side expression and right hand side expression both have same values. Therefore, this proves that $\left|\left(ka,kb\right)\right|=\left|k\right|\cdot \left|\left(a,b\right)\right|$ .