Analytic Geometry — Distance & Triangle Types (Part A)
Analytic Geometry — Distance & Triangle Types (Part A). Practice questions to deepen understanding of distance between points and types of triangles in analytic geometry. Online math practice with full solutions and step-by-step explanations.
Analytic Geometry — practice the distance formula between two points and identify isosceles and right triangles. Suitable for high school students at standard and advanced levels.
Find the distance between \((1, 2)\) and \((4, 6)\).
Using the formula: \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)\(d = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)
Find the distance between \((0, 0)\) and \((3, 4)\).
Right triangle with legs 3 and 4, so \(d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5\) This is the classic 3-4-5 triangle!
Find the distance between \((2, 3)\) and \((2, 8)\).
\(d = \sqrt{(2-2)^2 + (8-3)^2} = \sqrt{0 + 25} = 5\) The points share the same x, so the distance is the difference in y: |8-3| = 5
Find the distance between \((-1, -1)\) and \((2, 3)\).
\(d = \sqrt{(2-(-1))^2 + (3-(-1))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5\)
Find the distance between \((5, 1)\) and \((1, 4)\).
\(d = \sqrt{(1-5)^2 + (4-1)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\)
Find the distance between \((0, 3)\) and \((4, 0)\).
\(d = \sqrt{(4-0)^2 + (0-3)^2} = \sqrt{16 + 9} = 5\)
Find the distance between \((-2, 5)\) and \((3, -7)\).
\(d = \sqrt{(3-(-2))^2 + (-7-5)^2} = \sqrt{25 + 144} = \sqrt{169} = 13\) The 5-12-13 triangle!
Find the distance between \((1, 1)\) and \((4, 5)\).
\(d = \sqrt{(4-1)^2 + (5-1)^2} = \sqrt{9 + 16} = 5\)
Find the distance between \((6, 2)\) and \((2, 5)\).
\(d = \sqrt{(2-6)^2 + (5-2)^2} = \sqrt{16 + 9} = 5\)
Find the distance between \((-3, -4)\) and \((0, 0)\).
\(d = \sqrt{(0-(-3))^2 + (0-(-4))^2} = \sqrt{9 + 16} = 5\) The 3-4-5 triangle!
Find the distance between \((1, 2)\) and \((4, 6)\).
Using the formula: \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)\(d = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)
Find the distance between \((0, 0)\) and \((3, 4)\).
Right triangle with legs 3 and 4, so \(d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5\) This is the classic 3-4-5 triangle!
Find the distance between \((2, 3)\) and \((2, 8)\).
\(d = \sqrt{(2-2)^2 + (8-3)^2} = \sqrt{0 + 25} = 5\) The points share the same x, so the distance is the difference in y: |8-3| = 5
Find the distance between \((-1, -1)\) and \((2, 3)\).
\(d = \sqrt{(2-(-1))^2 + (3-(-1))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5\)
Find the distance between \((5, 1)\) and \((1, 4)\).
\(d = \sqrt{(1-5)^2 + (4-1)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\)
Find the distance between \((0, 3)\) and \((4, 0)\).
\(d = \sqrt{(4-0)^2 + (0-3)^2} = \sqrt{16 + 9} = 5\)
Find the distance between \((-2, 5)\) and \((3, -7)\).
\(d = \sqrt{(3-(-2))^2 + (-7-5)^2} = \sqrt{25 + 144} = \sqrt{169} = 13\) The 5-12-13 triangle!
Find the distance between \((1, 1)\) and \((4, 5)\).
\(d = \sqrt{(4-1)^2 + (5-1)^2} = \sqrt{9 + 16} = 5\)
Find the distance between \((6, 2)\) and \((2, 5)\).
\(d = \sqrt{(2-6)^2 + (5-2)^2} = \sqrt{16 + 9} = 5\)
Find the distance between \((-3, -4)\) and \((0, 0)\).
\(d = \sqrt{(0-(-3))^2 + (0-(-4))^2} = \sqrt{9 + 16} = 5\) The 3-4-5 triangle!
How many medians does a triangle have?
A triangle has 3 vertices, and from each vertex one median extends to the opposite side. Total: 3 medians.
Where do the medians of a triangle meet?
The three medians meet at a single point called the centroid. The centroid divides each median in a 2:1 ratio.
Where do the altitudes of a triangle meet?
The three altitudes meet at a single point called the orthocenter.
Where do the perpendicular bisectors of a triangle meet?
The three perpendicular bisectors meet at the circumcenter of the triangle.
The midsegment of a triangle is parallel to:
A midsegment is always parallel to the third side (the one it does not touch) and its length is half of that side.
The length of a midsegment in a triangle is:
The midsegment theorem: a segment connecting the midpoints of two sides is parallel to the third side and equals half its length.
In a triangle with vertices A(0,0), B(6,0), C(3,4), what is the midpoint of side AB?
Midpoint of AB is the average of the coordinates: ((0+6)/2, (0+0)/2) = (3,0)
The centroid divides each median in what ratio?
The centroid divides each median so that the part from the vertex to the centroid is twice the part from the centroid to the midpoint of the side.
In a right triangle, where is the orthocenter located?
In a right triangle, the two sides forming the right angle are altitudes. They meet at the vertex of the right angle.
In a triangle with A(0,4), B(4,0), C(4,4), what is the length of the midsegment parallel to AC?
The length of AC is 4 (straight horizontal). Midsegment = half = 2
Which line in a triangle is always inside the triangle?
A median is always inside the triangle. An altitude and perpendicular bisector can go outside in an obtuse triangle.
In a triangle with A(1,1), B(5,1), C(3,5), what is the centroid?
The centroid is the average of all coordinates: ((1+5+3)/3, (1+1+5)/3) = (3, 2.33 approximately)
Can a midsegment be longer than the side parallel to it?
The midsegment theorem states it is always half the parallel side, regardless of the type of triangle.
In an equilateral triangle, the median is also:
In an equilateral (and isosceles) triangle, the median to the base is also the altitude, angle bisector, and perpendicular bisector!
In a triangle with A(0,0), B(8,0), C(4,6), what is the length of the median from C to AB?
Midpoint of AB is (4,0). Distance from C(4,6) to (4,0) is |6-0| = 6
Find the area of a triangle with base 10 cm and height 6 cm.
Area of triangle: \(S = \frac{b \cdot h}{2} = \frac{10 \cdot 6}{2} = 30\) cm²
Find the area of a triangle with base 8 m and height 5 m.
Area: \(S = \frac{8 \cdot 5}{2} = 20\) m²
A triangle has area 24 cm² and base 8 cm. What is the height to that base?
From the formula: \(24 = \frac{8 \cdot h}{2}\)\(24 = 4h\) so \(h = 6\) cm
Find the area of a triangle with base 12 cm and height 7 cm.
Area: \(S = \frac{12 \cdot 7}{2} = 42\) cm²
A triangle has area 30 cm² and height 5 cm. What is the base length?
From the formula: \(30 = \frac{b \cdot 5}{2}\)\(60 = 5b\) so \(b = 12\) cm
Find the area of a right triangle with legs 3 cm and 4 cm.
In a right triangle the legs are the base and height: \(S = \frac{3 \cdot 4}{2} = 6\) cm²
A right triangle with legs 6 cm and 8 cm. What is its area?
A right triangle with legs 5 m and 12 m. What is the hypotenuse?
Pythagorean theorem: \(c = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\) meters
Egyptian triangle: 5, 12, 13
A right triangle with area 20 cm² and one leg 5 cm. What is the other leg?
From the formula: \(20 = \frac{5 \cdot b}{2}\)
\(40 = 5b\), therefore \(b = 8\) cm
A right triangle with legs 7 cm and 24 cm. What is the hypotenuse and area?
Hypotenuse: \(c = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = 25\)
Area: \(S = \frac{7 \cdot 24}{2} = 84\) square centimeters
In an isosceles triangle, the median to the base is also:
In an isosceles triangle, the median from the vertex to the base is also: an altitude (perpendicular to the base), an angle bisector (divides the vertex angle), and a perpendicular bisector of the base.
Isosceles triangle with legs 5 cm and base 6 cm. What is the height to the base?
The altitude divides the base into two equal parts, 3 cm each.
By Pythagoras: \(h = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = 4\) cm
Isosceles triangle with base 8 cm and height 3 cm. What is its area?
Area: \(S = \frac{8 \cdot 3}{2} = 12\) square centimeters
In an isosceles triangle, the base angles are:
In an isosceles triangle, the angles adjacent to the base are always equal to each other.
Isosceles triangle with legs 10 cm and base 12 cm. What is the height to the base?