Answer-
The corresponding height of the parallelogram is [tex]\dfrac{27}{5}[/tex] 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)
[tex]\Rightarrow \dfrac{y-y_1}{y_2-y_1}=\dfrac{x-x_1}{x_2-x_1}[/tex]
[tex]\Rightarrow \dfrac{y-5}{1-5}=\dfrac{x-4}{1-4}[/tex]
[tex]\Rightarrow \dfrac{y-5}{-4}=\dfrac{x-4}{-3}[/tex]
[tex]\Rightarrow \dfrac{y-5}{4}=\dfrac{x-4}{3}[/tex]
[tex]\Rightarrow 3(y-5)=4(x-4)[/tex]
[tex]\Rightarrow 3y-15=4x-16[/tex]
[tex]\Rightarrow 4x-3y-1=0[/tex]
Perpendicular distance between S and RU-
The distance d from a point (x₀, y₀) to the line ax+by+c=0 is
[tex]d=\dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}[/tex]
Distance of (7, 0) from line [tex]4x-3y-1=0[/tex] is,
[tex]d=\dfrac{|(4)(7)+(-3)(0)+(-1)|}{\sqrt{(4)^2+(-3)^2}}[/tex]
[tex]=\dfrac{|28-1|}{\sqrt{16+9}}[/tex]
[tex]=\dfrac{|27|}{\sqrt{25}}[/tex]
[tex]=\dfrac{27}{5}[/tex]
