(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 xintercept and the yintercept. Answer. To find the xintercept let y = 0 in y = x and solve for x. Thus, the xintercept is the point (, 0). To find the yintercept let x = 0 in y = x and solve for y. Thus, the yintercept is the point (0, ). University of Hawai i at Mānoa 1 R Spring  014
2 . Find the slopeintercept 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 pointslope equation of the line. Answer 5. y = (x ) (b) Write the equation in the slopeintercept form. Answer 6. (c) Find all intercepts. y = x + 5 Answer 7. The xintercept is the point (5, 0) and the yintercept 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 slopeintercept 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 slopeintercept form, we have y = x + so the slope is m =. The xintercept is the point (, 0) and the yintercept 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 pointslope 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 xintercept as the line L. Answer 1. We have m =, so m =. In slopeintercept 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 pointslope 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 pointslope 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 slopeintercept 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 yintercept. 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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