Question

Let y = [5 5] and u = [6 8] . Compute the distance from y to the line through u and the origin.

Answer 1

Answer:

The distance d = 1

Step-by-step explanation:

The objective is to compute the distance from y to the line through u and the origin.

Given that :

$y = \left[\begin{array}{cc}5\\5\end{array}\right]$  and $u = \left[\begin{array}{cc}6\\8\end{array}\right]$

Recall that:

from the  slope - intercept on the graph, the equation of line can be expressed as :

y = mx + b

where;

m = slope = $\dfrac{y_2-y_1}{x_2-x_1}$

b = y - intercept

Similarly, we are being informed that the line  passed through $u = \left[\begin{array}{cc}6\\8\end{array}\right]$ and origin, so ;

$x_1 = 0 , y_1 = 0 \\ \\ x_2 =6 , y_2 = 8$

the slope m = $\dfrac{y_2-y_1}{x_2-x_1}$

= $=\dfrac{8 - 0}{6-0}$

$= \dfrac{4}{3}$

Also, since the line pass through the origin:

Then

y = mx + b

0 = m(0) + b

b = 0

From y = mx + b

y = mx + (0)

y = mx

$y = \dfrac{4}{3}x$

3y = 4x

3y - 4x = 0

4x - 3y =0

The distance of a point (x,y) from a line ax +by + c = 0 can be represented with the equation:

$d = \dfrac{|ax+by +c|}{\sqrt{a^2 +b^2}}$

∴ the distance from $y = \left[\begin{array}{cc}5\\5\end{array}\right]$   to the line 4x - 3y = 0  is

$d = \dfrac{|4x-3y +0|}{\sqrt{4^2 +3^2}}$

$d = \dfrac{|4(5)-3(5) +0|}{\sqrt{16+9}}$

$d = \dfrac{20-15 }{\sqrt{25}}$

$d = \dfrac{5}{5}$

The distance d = 1

Answer 2

The distance from y(5,5) to the line through u(6,8) and the origin(0,0) is √2 units.

What is the distance between the two points on a graph?

The distance or length of any line on the graph is given by the formula,

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

where,

d = distance/length of the line between points 1 and 2,

(x₁, y₁) = coordinate of point 1,

(x₂, y₂) = coordinate of point 2,

In the given equation, we need to find the distance between the line and the point u (6,8). Now, we try to find the equation of the line, with points y=(5,5) and origin (0,0). Therefore, the slope of the equation can be written as,

$m = \dfrac{y_2-y_1}{x_2-x_1} = \dfrac{5-0}{5-0} = 1$

Now, if we substitute the value of the slope and a point in the general equation of the line, we will get,

$y = mx+c\\\\0 = 1(0)+c\\\\c = 0$

Further, if we draw the line on the graph, the nearest point to point u(6,8) is a(7,7). Therefore, the distance between the two points can be written as,

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\D = \sqrt{(7-6)^2+(7-8)^2}\\\\D = \sqrt2$

Hence, the distance from y(5,5) to the line through u(6,8) and the origin(0,0) is √2 units.

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