Question

The coordinates of the vertices of the triangle shown are p (2, 13), q (7, 1) and r (2, 1). what is the length of segment pq in units?

Answer 1

To find the length or distance of two coordinates, we use pythagoras theorem
l² = Δx² + Δy²

Because we want to know the length of PQ, we use coordinate P and coordinate Q
(xp, yp) = (2,13)
(xq, yq) = (7,1)

Input the coordinates
l² = Δx² + Δy²
l² = (xp - xq)² + (yp - yq)²
l² = (2-7)² + (13-1)²
l² = (-5)² + (12)²
l² = 25 + 144
l² = 169
l = √169
l = 13

The length is 13 unit length

Answer 2

The length of the segment PQ is 13 units.

Given that

The coordinates of the vertices of the triangle shown are p (2, 13), q (7, 1), and r (2, 1).

We have to determine

What is the length of segment PQ in units?

According to the question

The coordinates of the vertices of the triangle shown are p (2, 13), q (7, 1), and r (2, 1).

The length of the segment PQ is determined by the following formula;

$\rm PQ = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2$

The co-ordinate of triangle PQ is p(2, 13), q(7, 1).

Then,

The length of the segment PQ is;

$\rm PQ = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ \\ \rm PQ = \sqrt{(7-2)^2+(1-13)^2}\\ \\ \rm PQ = \sqrt{(5)^2+(-12)^2}\\ \\ \rm PQ = \sqrt{25+144}\\ \\ \rm PQ = \sqrt{169}\\ \\ \rm PQ = 13$

Hence, The length of the segment PQ is 13 units.

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