How to do coordinate geometry - HSPT Math

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Question

You are looking at a map of your town and your house is located at the coordinate (0,0). Your school is located at the point (3,4). If each coordinate distance is 1.3 miles, how far away is your school?

Answer

The coordinate length between you and your school is equivalent to the hypotenuse of a right triangle with sides of 3 and 4 units:

The distance is 5 coordinate lengths, and each coordinate length corresponds to 1.3 miles of distance, so

$5\times 1.3 =6.5\hspace{1 mm}miles$

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Question

Which of the following points will you find on the -axis?

Answer

A point is located on the -axis if and only if it has -coordinate (first coordinate) 0. Of the five choices, only fits that description.

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Question

Which of the following is a vertex of the square?

Answer

The coordinates of a point are determined by the distance from the origin. The first point in the ordered pair is the number of units to the left or right of the origin. Negative numbers indicate the number of units to the left while positive numbers indicate the number of units to the right. The second number indicates the number of units above or below the origin. Positive numbers indicate the number of units above while negative numbrs indicate the number of units below the origin. The vertices of the square are:
$\small (-1,2); (-1,4); (1,2); (-1,4)$

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Question

Which of the following points is on the -axis?

Answer

A point is located on the -axis if and only if it has a -coordinate equal to zero. So the answer is .

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Question

Billy set up a ramp for his toy cars. He did this by taking a wooden plank and putting one end on top of a brick that was 3 inches high. He then put the other end on top of a box that was 9 inches high. The bricks were 18 inches apart. What is the slope of the plank?

Answer

The value of the slope (m) is rise over run, and can be calculated with the formula below:

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

The coordinates of the first end of the plank would be (0,3), given that this is the starting point of the plank (so x would be 0), and y would be 3 since the brick is 3 inches tall.

The coordinates of the second end of the plank would be (18,9) since the plank is 18 inches long (so x would be 18) and y would be 9 since the box was 9 inches tall at the other end.

From this information we know that we can assign the following coordinates for the equation:

$\left(x_{1},y_{1}\right) = (0,3)$ and $(x_{2}, y_{2})= (18,9)$

Below is the solution we would get from plugging this information into the equation for slope:

$m = \frac{9-3}{18-0}}$

This reduces to $\frac{6}{18} =\frac{1}{3}$

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Question

A deer walks in a straight line for 8 hours. At the end of its journey, the deer is 30 miles north and 40 miles east of where it began. What was the average speed of the deer?

Answer

To find the speed of the deer, you must have the distance traveled and the time.

The distance is found using the Pythagorean Theorem:

$C = {\sqrt{A^2+ B^2}}$

$C = \sqrt{30^2 + 40^2}$

$C = \sqrt{2500} = 50$

The answer must be in miles per hour so the total miles are divided by the hours to get the final answer:

$\frac{miles}{hour} = \frac{50}{8} = 6.25 mph$

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Question

The point is reflected across . What is the new point?

Answer

The horizontal distance from point to the vertical line is two units. Since this point is reflected across , the new point will also be 2 units to the right of line .

Therefore, the correct answer is:

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Question

What is the slope given the following two points? $(2,3) \textup{ and } (3,-8)$

Answer

Write the slope formula and substitute the two points.

$m=\frac{y_2-y_1}{x_2-x_1}= \frac{-8-3}{3-2}= \frac{-11}{1} = -11$

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Question

Begin at the origin of the rectangular coordinate plane. Move up three units, left seven units, and down nine units. Give the coordinates of your current location.

Answer

Since you have moved left seven units, you have moved seven units in a negative horizontal direction, making the -coordinate of your current location .

Since you have moved up three units and down nine units, you have moved six units down - this is six units in a negative vertical direction, making the -coordinate of your current location .

Therefore, the ordered pair for your current location is .

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Question

The lines of which two of the following equations are perpendicular to each other?

(I) $y= \frac{6}{11} x$

(II) $y= - \frac{6}{11} x$

(III) $y= - \frac{11}{6} x$

Answer

All three equations are in slope-intercept form , so the slope of each line is the coefficient of :

(I) $y= \frac{6}{11} x$ - the slope is $\frac{6}{11}$

(II) $y= - \frac{6}{11} x$ - the slope is $- \frac{6}{11}$

(III) $y= - \frac{11}{6} x$ - the slope is $- \frac{11}{6}$

Two lines are perpendicular if and only if the product of their slopes is .

The product of the slopes of the lines in I and II is

$\frac{6}{11} \times \left (- \frac{6}{11} \right )= - \frac{6 \times 6 }{11 \times 11} =- \frac{36 }{121} e -1$

The product of the slopes of the lines in II and III is

$- \frac{6}{11} \times \left (- \frac{11} {6}\right )= + \frac{6 \times 11 }{11 \times 6} = \frac{66}{66} = 1 e -1$

The product of the slopes of the lines in I and III is

$\frac{6}{11} \times \left (- \frac{11} {6}\right )= - \frac{6 \times 11 }{11 \times 6} =- \frac{66}{66} = -1$

This makes the lines in I and III perpendicular, and this is the correct choice.

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Question

What is the slope of a line through the points and ?

Answer

Use the slope fomula, setting $x_{1} = -2, x_{2} = 2,y_{1} = 3, y_{2} = 5$

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{5-3}{2-(-2)} = \frac{2}{4} =\frac{1}{2}$

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Question

In which quadrant, or on which axis, is the point with coordinates located?

Answer

Any point with a positive -coordinate and a negative -coordinate is located in the lower right quadrant - Quadrant IV.

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Question

Which of the following is an equation of a line with slope 0?

Answer

A line with slope 0 has equation for some real value .

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Question

Give the equation of the red line in slope-intercept form.

Answer

The slope of the line is

$m = \frac{y_{2}- y_{1}}{x_{2}- x_{1}} = \frac{6- (-1)}{0-(-4)} = \frac{7}{4}$

The -intercept of the line has -coordinate

The slope-intercept form can be written:

Replace:

$y = \frac{7}{4}x+6$

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Question

The green and blue lines are perpendicular: What is the slope of the blue line?

Answer

The slope of the blue line, being perpendicular to the green line, is the opposite of the reciprocal of the slope of the green line. The slope of the green line can be found using the slope formula:

$m = \frac{y_{2}- y_{1}}{x_{2}- x_{1}} = \frac{10-0}{0-30} = -\frac{1}{3}$

The opposite of the reciprocal of $-\frac{1}{3}$ is 3, and this is the slope of the blue line.

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Question

A line segment on the coordinate plane has endpoints and . In terms of and , as applicable, give the -coordinate of its midpoint.

Answer

The -coordinate of the midpoint of a line segment is the mean of the -coordinates of its endpoints. Therefore, the -coordinate is

$y= \frac{(a-4)+(-b)}{2} = \frac{a-b-4}{2}$ .

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Question

A line segment on the coordinate plane has endpoints and . In terms of and , as applicable, give the -coordinate of its midpoint.

Answer

The -coordinate of the midpoint of a line segment is the mean of the -coordinates of its endpoints. Therefore, the -coordinate is

$x= \frac{(a-5)+(b+8)}{2} = \frac{a+b-5+8}{2} = \frac{a+b+3}{2}$ .

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Question

What is the -intercept of the graph of the function

$f(x) = x^{2} + 3x - 10?$

Answer

The -intercept of the graph of a function is the point at which it intersects the -axis - that is, at which . This point is $\left (0, f(0) \right )$ , so evaluate :

$f(x) = x^{2} + 3x - 10$

$f( 0) = 0^{2} + 3 \cdot 0 - 10 = 0 + 0 - 10 = -10$

The -intercept is $\left (0, -10 \right )$ .

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Question

A line segment on the coordinate plane has one endpoint at $\left ( 8, 10 \right )$ ; its midpoint is $\left ( A+B, A- B\right )$ . Which of the following gives the -coordinate of its other endpoint in terms of and ?

Answer

To find the value of the -coordinate of the other endpoint, we will assign the variable . Then, since the -coordinate of the midpoint of the segment is the mean of those of its endpoints, the equation that we can set up is

$\frac{T + 8 }{2} = A + B$ .

We solve for :

$\frac{T + 8 }{2} \cdot 2 =\left ( A + B \right )\cdot 2$

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Question

Two perpendicular lines intersect at the point . One line passes through point ; the other passes through point $\left ( 4, Y\right )$ . Evaluate .

Answer

The line that passes through $\left ( 3,1\right )$ and $\left ( 5,4\right )$ has slope

$m = \frac{4-1}{5-3} = \frac{3}{2}$ .

The line that passes through $\left ( 4, Y\right )$ and , being perpendicular to the first, has as its slope the opposite reciprocal of $\frac{3}{2}$ , or $-\frac{2}{3}$ .

Therefore, to find , we use the slope formula and solve for :

$-\frac{2}{3} = \frac{Y-4}{4-5}$

$-\frac{2}{3} = \frac{Y-4}{-1}$

$-\frac{2}{3} \cdot (-1) = \frac{Y-4}{-1} \cdot (-1)$

$\frac{2}{3}= Y-4$

$\frac{2}{3}+ 4= Y-4 + 4$

$4\frac{2}{3}= Y$

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