Problem of The Day A vector v of length 6 makes a 150 degree angle with the vector [1,0], when they are placed tail-to-tail. Find the components of v.

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Presentation transcript:

Problem of The Day A vector v of length 6 makes a 150 degree angle with the vector [1,0], when they are placed tail-to-tail. Find the components of v.

Distance between a point and a line Objective: To be able to find the distance between a point and a line TS: Making decisions after reflection and review Warm up: Write y= (1/2)x - (3/2) in general form

To find the distance between a point (x1, y1) and a line whose general form is Ax + By + C = 0 (where A, B & C must be integers), you can use the formula where d is the shortest (perpendicular) distance between the point and the line. For example: To find the distance between a point (2, 4) and the line y = ½ x – 3/2, I would first get the line into general form: 2y = x – 3 0 = x – 2y – 3 So A = 1, B = -2 and C = -3 Now I would plug into the formula to find the distance between the point and the line This means your first step is to identify a point you’re measuring from and the equation (in general form, Ax + By + C = 0) of the line you’re measuring to.

Find the distance from the point to the line: 1. (4, -1) and 4x + 3y – 6 = 0 2. (1, 5) and 5x – 12y + 7 = 0 3. (-1, 2) and 3x + y + 8 = 0 4. (0, 3) and -2x + 5y – 6 = 0

5. 4x – 5y + 8 = 0 6. 2x + 7y + 5 = 0 4x – 5y + 2 = 0 2x + 7y – 3 = 0 Find the distance between the parallel lines: 5. 4x – 5y + 8 = 0 6. 2x + 7y + 5 = 0 4x – 5y + 2 = 0 2x + 7y – 3 = 0

A line is parallel to 3x – 4y = 6, and 3 units from the origin A line is parallel to 3x – 4y = 6, and 3 units from the origin. Find the equation of the line.

A line is parallel to the line through A(-6, 2) and B(6, 7) and is tangent to the circle with center at the origin and radius of 5. Find the equation of the line.

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In triangle ABC, find the lengths of the three altitudes In triangle ABC, find the lengths of the three altitudes. A(-5, -2) B(1, 6) C(7, 3)