Can You Pass The Coordinate Geometry Quiz?

1) If (4,1) is the midpoint of the interval from (x,-2) to (5,y), what is the value of y?

0

6

-3

-10

4

The midpoint formula states that the midpoint of a line segment is the average of the x-coordinates and the average of the y-coordinates of the endpoints. In this case, the x-coordinate of the midpoint is 4, which is the average of x and 5. Therefore, x = 3. The y-coordinate of the midpoint is 1, which is the average of -2 and y. Therefore, -2 + y = 2, and solving for y gives y = 4.

Explanation

The midpoint formula states that the midpoint of a line segment is the average of the x-coordinates and the average of the y-coordinates of the endpoints. In this case, the x-coordinate of the midpoint is 4, which is the average of x and 5. Therefore, x = 3. The y-coordinate of the midpoint is 1, which is the average of -2 and y. Therefore, -2 + y = 2, and solving for y gives y = 4.

2) The line through the point (-3,-5) parallel to the x-axis has the equation.

X=-3

X=-5

Y=-3

Y=-5

5x=3y

The correct answer is y=-5 because a line parallel to the x-axis has a constant y-value. Since the line passes through the point (-3,-5), the y-value for any point on the line will always be -5. Therefore, the equation of the line is y=-5.

Explanation

The correct answer is y=-5 because a line parallel to the x-axis has a constant y-value. Since the line passes through the point (-3,-5), the y-value for any point on the line will always be -5. Therefore, the equation of the line is y=-5.

3) If the point (x,y) is first reflected in the line x=0 and the resulting point is then reflected in the line y=0, then the image point has coordinates

(x,1-y)

(0,1)

(-x,1-y)

(-x, -y)

(y,-x)

When a point is reflected in the line x=0, the x-coordinate remains the same, but the sign of the y-coordinate is reversed. So, the new point after the first reflection would be (x,-y).

When this new point is reflected in the line y=0, the y-coordinate remains the same, but the sign of the x-coordinate is reversed. So, the final image point would be (-x, -y).

Explanation

When a point is reflected in the line x=0, the x-coordinate remains the same, but the sign of the y-coordinate is reversed. So, the new point after the first reflection would be (x,-y).

When this new point is reflected in the line y=0, the y-coordinate remains the same, but the sign of the x-coordinate is reversed. So, the final image point would be (-x, -y).

4) The gradient of a straight line is -3/2 and it cuts the x-axis at the point (4,0). The equation of the line is

2y-3x=6

2y=-3x+12

2y=3x+8

2y=3x-12

2y=3x-8

The equation of a straight line can be written in the form y = mx + c, where m is the gradient and c is the y-intercept. In this case, the gradient is -3/2, which means that for every 2 units the line moves horizontally (x-axis), it moves downward 3 units (y-axis).

Since the line cuts the x-axis at the point (4,0), we know that when x = 4, y = 0. Plugging these values into the equation y = mx + c, we get 0 = (-3/2)(4) + c. Simplifying this equation gives us c = 6.

Therefore, the equation of the line is 2y = -3x + 12.

Explanation

The equation of a straight line can be written in the form y = mx + c, where m is the gradient and c is the y-intercept. In this case, the gradient is -3/2, which means that for every 2 units the line moves horizontally (x-axis), it moves downward 3 units (y-axis).

Since the line cuts the x-axis at the point (4,0), we know that when x = 4, y = 0. Plugging these values into the equation y = mx + c, we get 0 = (-3/2)(4) + c. Simplifying this equation gives us c = 6.

Therefore, the equation of the line is 2y = -3x + 12.

5) The point (-2,4) is the mid point of the line segment PQ where P is the point (2,-2). The coordinates of Q are:

(0,1)

(-6,6)

(6,-6)

(-2,6)

(-6,10)

The midpoint of a line segment is the average of the coordinates of its endpoints. Given that the midpoint is (-2,4) and one endpoint is (2,-2), we can find the coordinates of the other endpoint by subtracting the coordinates of the midpoint from twice the coordinates of one endpoint. Using this formula, we get Q = 2P - (-2,4) = (2*2 - (-2), 2*(-2) - 4) = (-6,10). Therefore, the coordinates of Q are (-6,10).

Explanation

The midpoint of a line segment is the average of the coordinates of its endpoints. Given that the midpoint is (-2,4) and one endpoint is (2,-2), we can find the coordinates of the other endpoint by subtracting the coordinates of the midpoint from twice the coordinates of one endpoint. Using this formula, we get Q = 2P - (-2,4) = (2*2 - (-2), 2*(-2) - 4) = (-6,10). Therefore, the coordinates of Q are (-6,10).

6) The equation of the straight line passing through (3,5) perpendicular to 3x+y=6 is:

3y+x=6

3y-x-12=0

3y+x=18

3y+x+6=0

3y-x-18=0

The equation of a straight line passing through a point (3,5) perpendicular to another line can be found by taking the negative reciprocal of the slope of the given line. The given line, 3x+y=6, can be rewritten as y=-3x+6, which has a slope of -3. The negative reciprocal of -3 is 1/3. Using the point-slope form of a line, y-y1=m(x-x1), where (x1,y1) is the given point and m is the slope, we can substitute (3,5) and 1/3 into the equation. Simplifying, we get 3y-x-12=0, which matches the given answer.

Explanation

The equation of a straight line passing through a point (3,5) perpendicular to another line can be found by taking the negative reciprocal of the slope of the given line. The given line, 3x+y=6, can be rewritten as y=-3x+6, which has a slope of -3. The negative reciprocal of -3 is 1/3. Using the point-slope form of a line, y-y1=m(x-x1), where (x1,y1) is the given point and m is the slope, we can substitute (3,5) and 1/3 into the equation. Simplifying, we get 3y-x-12=0, which matches the given answer.

7) Given 3 points, P, Q, and R with R on the x-axis, the equation of RQ is y=2x-1. if QP is parallel to the x-axis and the coordinates of P is (8,4), then the distance from P to Q is:

3.5

4

4.5

5

5.5

Since QP is parallel to the x-axis, the y-coordinate of Q will be the same as the y-coordinate of P, which is 4. We can substitute this value into the equation of RQ to find the x-coordinate of Q. By rearranging the equation, we get 2x = y + 1, so 2x = 4 + 1, and x = 2.5. Therefore, the coordinates of Q are (2.5, 4). To find the distance between P and Q, we can use the distance formula: √((x2 - x1)^2 + (y2 - y1)^2). Plugging in the coordinates, we get √((2.5 - 8)^2 + (4 - 4)^2) = √((-5.5)^2 + 0^2) = √(30.25) = 5.5. Therefore, the distance from P to Q is 5.5.

Explanation

Since QP is parallel to the x-axis, the y-coordinate of Q will be the same as the y-coordinate of P, which is 4. We can substitute this value into the equation of RQ to find the x-coordinate of Q. By rearranging the equation, we get 2x = y + 1, so 2x = 4 + 1, and x = 2.5. Therefore, the coordinates of Q are (2.5, 4). To find the distance between P and Q, we can use the distance formula: √((x2 - x1)^2 + (y2 - y1)^2). Plugging in the coordinates, we get √((2.5 - 8)^2 + (4 - 4)^2) = √((-5.5)^2 + 0^2) = √(30.25) = 5.5. Therefore, the distance from P to Q is 5.5.

8) The x-coordinate of the foot of the perpendicular from the point(1,9) to the line y=x is

4

5

6

7

9

The x-coordinate of the foot of the perpendicular from the point (1,9) to the line y=x can be found by finding the equation of the line perpendicular to y=x passing through the point (1,9) and then finding the x-coordinate of the intersection point of the two lines. The equation of the line perpendicular to y=x can be found by taking the negative reciprocal of the slope of y=x, which is -1. Using the point-slope form of a line, the equation of the perpendicular line is y - 9 = -1(x - 1). Simplifying this equation gives y = -x + 10. To find the x-coordinate of the intersection point, we set y = x and solve for x. Thus, x = 5.

Explanation

The x-coordinate of the foot of the perpendicular from the point (1,9) to the line y=x can be found by finding the equation of the line perpendicular to y=x passing through the point (1,9) and then finding the x-coordinate of the intersection point of the two lines. The equation of the line perpendicular to y=x can be found by taking the negative reciprocal of the slope of y=x, which is -1. Using the point-slope form of a line, the equation of the perpendicular line is y - 9 = -1(x - 1). Simplifying this equation gives y = -x + 10. To find the x-coordinate of the intersection point, we set y = x and solve for x. Thus, x = 5.

9) If the line y=2x+3 is reflected in the line y=x+1, its equation becomes

2x-y=0

X-2y+1=0

X-2y+0

X-2y-2=0

2x-2y-2+0

When a line is reflected in another line, the equation of the reflected line is obtained by replacing x with y and y with x in the original equation. In this case, the original equation is y=2x+3. Replacing x with y and y with x, we get x=2y+3. Simplifying this equation, we get x-2y-3=0. However, none of the given options match this equation exactly. The closest option is x-2y+0, which can be simplified to x-2y=0. Therefore, x-2y+0 is the closest approximation to the equation of the reflected line.

Explanation

When a line is reflected in another line, the equation of the reflected line is obtained by replacing x with y and y with x in the original equation. In this case, the original equation is y=2x+3. Replacing x with y and y with x, we get x=2y+3. Simplifying this equation, we get x-2y-3=0. However, none of the given options match this equation exactly. The closest option is x-2y+0, which can be simplified to x-2y=0. Therefore, x-2y+0 is the closest approximation to the equation of the reflected line.

10) The points (-1,6), (0,0) and (3,1) are three vertices of a parallelogram. The number of possible positions of the fourth vertex is

0

1

2

3

4

Given three points (-1,6), (0,0), and (3,1), we can determine the fourth vertex of the parallelogram by finding the midpoint of the line segment connecting the first two points and then extending that line segment by the same length to find the fourth point. Similarly, we can find the midpoint of the line segment connecting the second and third points and extend it to find the fourth point. Finally, we can find the midpoint of the line segment connecting the third and first points and extend it to find the fourth point. Therefore, there are three possible positions for the fourth vertex of the parallelogram.

Explanation

Given three points (-1,6), (0,0), and (3,1), we can determine the fourth vertex of the parallelogram by finding the midpoint of the line segment connecting the first two points and then extending that line segment by the same length to find the fourth point. Similarly, we can find the midpoint of the line segment connecting the second and third points and extend it to find the fourth point. Finally, we can find the midpoint of the line segment connecting the third and first points and extend it to find the fourth point. Therefore, there are three possible positions for the fourth vertex of the parallelogram.