Question

If line segment RU is considered the base of parallelogram RSTU, what is the corresponding height of the parallelogram?

Answer 1

Vertices: (1,1) ,(4,5) , (10,4) , (7,0)
we can use the cross product 
| RU X RS |  / distance(1,1,4,5)

b)5.4 units

Answer 2

Answer-

The corresponding height of the parallelogram is $\dfrac{27}{5}$ units

Solution-

Hint- The perpendicular distance between the point S and the straight line RU is the length of the height of the parallelogram.

Equation of RU-

Applying two point formula between (4, 5), (1, 1)

$\Rightarrow \dfrac{y-y_1}{y_2-y_1}=\dfrac{x-x_1}{x_2-x_1}$

$\Rightarrow \dfrac{y-5}{1-5}=\dfrac{x-4}{1-4}$

$\Rightarrow \dfrac{y-5}{-4}=\dfrac{x-4}{-3}$

$\Rightarrow \dfrac{y-5}{4}=\dfrac{x-4}{3}$

$\Rightarrow 3(y-5)=4(x-4)$

$\Rightarrow 3y-15=4x-16$

$\Rightarrow 4x-3y-1=0$

Perpendicular distance between S and RU-

The distance d from a point (x₀, y₀) to the line ax+by+c=0 is

$d=\dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$

Distance of (7, 0) from line $4x-3y-1=0$ is,

$d=\dfrac{|(4)(7)+(-3)(0)+(-1)|}{\sqrt{(4)^2+(-3)^2}}$

$=\dfrac{|28-1|}{\sqrt{16+9}}$

$=\dfrac{|27|}{\sqrt{25}}$

$=\dfrac{27}{5}$