Tell whether the lines through the given points are parallel, perpendicular, or neither.

Mathematics

1 answer:

zubka84 [21]3 years ago

7 0

line 1: slope=2/3

slope: (4-0) / (7-1)

4/6 = 2/3

line 2: slope= 6/-4

slope: (6-0)/(3-7)

6/-4

These two lines are perpendicular

You might be interested in

Help plz with 17,18,19

Brilliant_brown [7]

17 is C, 18 is J,and 19 is B.We just and a math test in this and I got an A+

3 0

3 years ago

Read 2 more answers

Given: △ABC, m∠A=60°, m∠C=45°, AB=9 Find: Perimeter of △ABC, Area of △ABC

stellarik [79]

Answer:

Perimeter of ΔABC: $\frac{27}{2}$ + $\frac{9}{2} * \sqrt6$ units

Area of ΔABC: $\frac{81}{8}*\sqrt3 + \frac{243}{8}$ units

Skills required: HS Geo, Special Triangles

Step-by-step explanation:

1) The best option is to break down this triangle. Let's draw an altitude from Point B down to Segment AC. The point from the altitude that intersects AC is Point D. BD is the height of our triangle, AC is the base.

2) Angle A is 60 degrees, and since Angle BDA is 90 degrees, Angle ABD is 30 degrees. We can use the 30-60-90 degree right triangle property for the triangle BDA.

This states that if the side opposite the 30 degree angle is $x$ , the side opposite the 60 degree angle is $x*\sqrt3$ , and the side opposite the 90 degree angle is $2x$ .

AB is 9 units, and it is opposite the 90 degree angle. This means that $2x=9, x = \frac{9}{2}$ ==> This then means that AD, the segment opposite the 30 degree angle in this triangle is $\frac{9}{2}$ units. Segment BD (the height) is $\frac{9}{2} * \sqrt3$ .

3) Angle C is 45 degrees, and Angle BDC is 90 degrees, which means that Angle CBD is 45 degrees. We can use the 45-45-90 degree right triangle property for the triangle BCD.

This states that if the side opposite the 45 degree angle is $x$ , the other side opposite a 45 degree angle is also $x$ , but the hypotenuse (side opposite the right (90 degree) angle) is $\sqrt{2}*x$ .

BD is $\frac{9}{2} * \sqrt3$ , which means DC is the same. BC, which is the hypotenuse is BD multiplied by square-root-2, which is $\frac{9}{2} * \sqrt6$ .

4) Area is $\frac{1}{2}*b*h$ , the base (b) is AC (which is $\frac{9}{2}+\frac{9}{2}*\sqrt3$ ), the height is BD ( $\frac{9}{2}*\sqrt3$ ). When multiple you will get $\frac{81}{4}*\sqrt3 + \frac{243}{4}$ , then this multiplied by 1/2 is