(a) Find five points on the line and arrange them in a table. Answer 1. y = 3x 2 x y

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1 1. Given is the line with equation y = x. (a) Find five points on the line and arrange them in a table. Answer 1. (b) Graph the line. Answer. y = x x y y y = x (1, 1) x (0, ) (c) Find the x-intercept and the y-intercept. Answer. To find the x-intercept let y = 0 in y = x and solve for x. Thus, the x-intercept is the point (, 0). To find the y-intercept let x = 0 in y = x and solve for y. Thus, the y-intercept is the point (0, ). University of Hawai i at Mānoa 1 R Spring - 014

2 . Find the slope-intercept form of the equation of the line through the points (, 7) and (5, ) and graph it. Answer 4. First calculate the slope. m = 7 = 5. So far, we have 5 y = 5 x + b. To solve for b we substitute the coordinates of a point on the line, for example (, 7). Then at the point (, 7), we have 7 = 5 + b b = b = 1, and so the answer is y = 5 x + 1. We should check our work by verifying that the other point also lies on the line. In other words, substituting the point (5, ) we should obtain an identity. Indeed, = = = 6 =.. Consider the line passing through the point (, ) with slope m = 1. (a) Write down the point-slope equation of the line. Answer 5. y = (x ) (b) Write the equation in the slope-intercept form. Answer 6. (c) Find all intercepts. y = x + 5 Answer 7. The x-intercept is the point (5, 0) and the y-intercept is the point (0, 5). University of Hawai i at Mānoa 1 R Spring - 014

3 4. Consider the line y = x +. (a) Find the equation in slope-intercept form of a parallel line through (, 5). Answer 8. The given line has slope m =, so we are looking for a line of the form y = x + b and containing the point (, 5). Substituting x =, it follows that b = 1 in order for y = 5. Thus, we obtain y = x + 1. (b) Find the equation of a perpendicular line through (, 7). Answer 9. The line has slope m =, so m = 1 = 1, and a perpendicular m line will have the form y = 1 x + b. Substituting the point (, 7) and solving for b, we obtain 5. Consider the line L given by x + y = 6. y = 1 x + 8. (a) Find the slope and intercepts of the line. Answer 10. In the slope-intercept form, we have y = x + so the slope is m =. The x-intercept is the point (, 0) and the y-intercept is the point (0, ). (b) Find a point on the line and a point not on the line. Answer 11. The point (0, 0) does not lie on the line, but (, 0) does. (c) Write the equation of the line in point-slope form. Answer 1. The slope we already know to be m = and we can choose the point (, 0), so y = (x ). (d) Find the equation of a line perpendicular to L, but passing through the same x-intercept as the line L. Answer 1. We have m =, so m =. In slope-intercept form, we have y = x + b, and we need to have this line pass through the point (, 0). Substituting we find that b = 9, and so the answer is y = x Solve: { y = x 1 x 5y = 10 University of Hawai i at Mānoa 14 R Spring - 014

4 Answer 14. Proceed by elimination: rewrite the system of equations and add them. We have, { x + y = 1 x 5y = 10, and whence 4y = 9 y = 9. Then we substitute y = 9 into the first equation 4 4 and solve for x and obtain x = 5. To make sure that ( 5, 9 ) is the solution we check that it also solves the second equation, ( 5 ) ( 5 9 ) = = 40 4 = Derive the point-slope form of the equation for a line by following these steps. (a) Let L be the line passing through the fixed point (x 1, y 1 ) and an arbitrary point (x, y). (b) Find the general formula for the slope of L. Answer 15. The slope of L is given by m = y y 1 x x 1. Multiplying thru by (x x 1 ), we obtain the point-slope form. 8. *Write down a system of linear equations that has (a) Exactly one solution. Answer 16. All the above problems have exactly one solution. Take, for example, Problem 6 and introduce a third line which passes through the solution ( 5, 9 ). We use the slope-intercept form with an arbitrary slope, say m =, 8 4 and obtain y = x 1 y + 9 = (x + 5 ). 4 8 x 5y = 10 (b) No solution. Answer 17. The only three lines in the plane that do not intersect are parallel lines. We can take for example the line x 5y = 10 and pick different y- intercepts. x 5y = 0 x 5y = 5 x 5y = 10. (c) Infinitely many solutions. Answer 18. Infinitely many solutions occur when the three lines are in fact the same line. That is, we have three parallel lines with the same y-intercept. x 5y = 0 4x 10y = 0 = 0 University of Hawai i at Mānoa 15 R Spring - 014

5 9. **Find a pair of points that together with the points (, 1) and (, ) are the vertices of a square. Answer 19. Case 1: The segment (, 1)(, ) is a side of a square. The line through the points (, 1) and (, ) has the equation and the distance between these points is y + = 4 (x ), d = ( 1) + ( + ) = 5. Each of the two points we are looking for needs to lie on a line parallel to the one above and also on a line perpendicular to it and passing thru either (, 1) or (, ). That is, we need to solve the systems { y + = 4 (x ) (y + ) + (x ) = 5, { and y 1 = 4 (x + ) (y 1) + (x + ) = 5. In the first system, we substitute for (y + ) in the second equation. (y + ) + (x ) = 5 (1) (4 ) (x ) + (x ) = 5 () 16 9 (x ) + (x ) = 5 () 5 9 (x ) = 5 (4) 5 9 (x ) = 5 (5) (x ) = 9 (6) x 4x 5 = 0 (7) (x 5)(x + 1) = 0 (8) Thus we obtain the solutions (5, ) and ( 1, 6). Following the same procedure for the second system, we obtain the solutions ( 5, ) and (1, 5). But our pair of solutions must lie on a line parallel to the one thru (, 1) and (, ); i.e., a line with slope m =. So, the possible solutions are the pairs of points (1, 5), (5, ), 4 and ( 5, ), ( 1, 6). University of Hawai i at Mānoa 16 R Spring - 014

6 Case : The segment (, 1)(, ) lies on the diagonal of a square. In this case we will obtain a smaller square of side length 5. We need to find the other diagonal; i.e., the line perpendicular to the segment (, 1)(, ) and passing through its midpoint ( +, 1 ) = (0, 1 ). This line has the equation y + 1 = 4 x, and since half the diagonal is 5, the two points we are looking for need to be distance 5 away from the center of the square, the point (0, 1 ), as well as the endpoints of (, 1)(, ). We need to solve the system { y + 1 = 4 x (y + 1 ) + (x ) = ( 5 ). Omitting the algebra, we obtain the points (, ) and (, 5 ), and this is the third possible solution. University of Hawai i at Mānoa 17 R Spring - 014

7 10. ***Find all points such that together with the points (, 1) and (, ) they are the vertices of a right triangle. Answer 0. We have two cases to consider. First, suppose that the line segment (, 1)(, ) is the leg of a right triangle. Then, the third vertex lies on a line perpendicular to the line thru (, 1) and (, ), and passing thru either (, 1) or (, ). If (x, y ) is the third vertex, then either y + = 4 (x ) or y 1 = 4 (x + ). The second and more interesting case is that the line segment (, 1)(, ) is the hypotenuse of a right triangle. From elementary geometry we recall the Theorem of Thales, which states that the triangle formed by the diameter of a circle and line segments joining an arbitrary point on the circle with the endpoints of the diameter is a right triangle. Thus, we need to find the equation of a circle whose diameter is the line segment (, 1)(, ). Using the distance formula, we have the diameter d = ( 1) + ( + ) = 5, and the midpoint of our circle, ( x1 + x, y ) ( 1 + y = 0, 1 ) This is the circle of radius 5, centered at the point (0, 1 ). The answer is disappointing, because we do not explicitly give the coordinates of a point or points. In fact there are infinitely many possibilities, so listing them amounts to writing a formula which computes them for us. We have a formula for the coordinates of every point which solves the problem. If (x, y ) is the third vertex of the right triangle with vertices (, 1) and (, ), then either (x ) + y + = 4 (x ), or y 1 = 4 (x + ), or ( y + 1 ) = ( ) 5.. University of Hawai i at Mānoa 18 R Spring - 014

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