How to find out if a point is on a line with an equation - SAT Math

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Question

In the xy -plane, line l is given by the equation 2_x_ - 3_y_ = 5. If line l passes through the point (a ,1), what is the value of a ?

Answer

The equation of line l relates x -values and y -values that lie along the line. The question is asking for the x -value of a point on the line whose y -value is 1, so we are looking for the x -value on the line when the y-value is 1. In the equation of the line, plug 1 in for y and solve for x:

2_x_ - 3(1) = 5

2_x_ - 3 = 5

2_x_ = 8

x = 4. So the missing x-value on line l is 4.

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Question

For the line

$y=\frac{1}{3}x-7$

Which one of these coordinates can be found on the line?

Answer

To test the coordinates, plug the x-coordinate into the line equation and solve for y.

y = 1/3x -7

Test (3,-6)

y = 1/3(3) – 7 = 1 – 7 = -6 YES!

Test (3,7)

y = 1/3(3) – 7 = 1 – 7 = -6 NO

Test (6,-12)

y = 1/3(6) – 7 = 2 – 7 = -5 NO

Test (6,5)

y = 1/3(6) – 7 = 2 – 7 = -5 NO

Test (9,5)

y = 1/3(9) – 7 = 3 – 7 = -4 NO

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Question

Consider the lines described by the following two equations:

4y = 3x2

3y = 4x2

Find the vertical distance between the two lines at the points where x = 6.

Answer

Since the vertical coordinates of each point are given by y, solve each equation for y and plug in 6 for x, as follows:

Taking the difference of the resulting y -values give the vertical distance between the points (6,27) and (6,48), which is 21.

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Question

The equation of a line is: 2x + 9y = 71

Which of these points is on that line?

Answer

Test the difference combinations out starting with the most repeated number. In this case, y = 7 appears most often in the answers. Plug in y=7 and solve for x. If the answer does not appear on the list, solve for the next most common coordinate.

2(x) + 9(7) = 71

2x + 63 = 71

2x = 8

x = 4

Therefore the answer is (4, 7)

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Question

Solve the following system of equations:

–2x + 3y = 10

2x + 5y = 6

Answer

Since we have –2x and +2x in the equations, it makes sense to add the equations together to give 8y = 16 yielding y = 2. Then we substitute y = 2 into one of the original equations to get x = –2. So the solution to the system of equations is (–2, 2)

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Question

$\dpi{100} \small 5x+25y = 125$

Which point lies on this line?

Answer

$\dpi{100} \small 5x+25y = 125$

Test the coordinates to find the ordered pair that makes the equation of the line true:

$\dpi{100} \small (5,4)$

$\dpi{100} \small 5 (5) + 25 (4) = 25 + 100 = 125$

$\dpi{100} \small (1,5)$

$\dpi{100} \small 5(1)+25(5)= 5+125=130$

$\dpi{100} \small (5,1)$

$\dpi{100} \small 5(5)+25(1)= 25+25=50$

$\dpi{100} \small (5,5)$

$\dpi{100} \small 5(5)+25(5)= 25+125=150$

$\dpi{100} \small (1,4)$

$\dpi{100} \small 5(1)+25(4)= 5+100=105$

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Question

Which of the following lines contains the point (8, 9)?

Answer

In order to find out which of these lines is correct, we simply plug in the values $\dpi{100} \small x=8$ and $\dpi{100} \small y=9$ into each equation and see if it balances.

The only one for which this will work is $\dpi{100} \small 3x-6=2y$

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Question

Which of the following sets of coordinates are on the line $y=3x-4$ ?

Answer

$(2,2)$ when plugged in for $y$ and $x$ make the linear equation true, therefore those coordinates fall on that line.

$y=3x-4$

Because this equation is true, the point must lie on the line. The other given answer choices do not result in true equalities.

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Question

Which of the following points can be found on the line $\small y=3x+2$ ?

Answer

We are looking for an ordered pair that makes the given equation true. To solve, plug in the various answer choices to find the true equality.

Because this equality is true, we can conclude that the point lies on this line. None of the other given answer options will result in a true equality.

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Question

Which of the following points is not on the line $y=\frac{5}{2}x+3$ ?

Answer

To figure out if any of the points are on the line, substitute the and coordinates into the equation. If the equation is incorrect, the point is not on the line. For the point :

$12.5 eq \frac{5}{2}(4)+3$

So, is not on the line.

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Question

At what point do these two lines intersect?

Answer

If two lines intersect, that means that at one point, the and values are the same. Therefore, we can use substitution to solve this problem.

Let's substitute in for in the other equation. Then, solve for :

Now, we can substitute this into either equation and solve for :

With these two values, the point of intersection is

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Question

At what point do these two lines intersect?

Answer

If two lines intersect, that means that their and values are the same at one point. Therefore, we can use substitution to solve this problem.

First, let's write these two formulas in slope-intercept form. First:

$y=\frac{2}{3}x - \frac{1}{3}$

Then, for the second line:

$y=-\frac{3}{2}x +1$

Now, we can substitute $y=\frac{2}{3}x - \frac{1}{3}$ in for in our second equation and solve for , like so:

$\frac{2}{3}x -\frac{1}{3} = -\frac{3}{2}x +1$

$\frac{2}{3}x +\frac{3}{2}x = \frac{4}{3}$

$\frac{4}{6}x +\frac{9}{6}x = \frac{4}{3}$

$\frac{15}{6}x = \frac{4}{3}$

$x = \frac{24}{45} = \frac{8}{15}$

Now, we can substitute this value into either equation to solve for .

$2(\frac{8}{15}) - 3y =1$

$\frac{16}{15} - 3y =1$

$-3y = 1-\frac{18}{15}$

$y = -3(\frac{15}{15} - \frac{16}{15}) = -3(-\frac{1}{15}) = \frac{3}{15}$

Therefore, our point of intersection is $(\frac{8}{15}, \frac{3}{15})$

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Question

Figure NOT drawn to scale.

On the coordinate axes shown above, the shaded triangle has the following area:

Evaluate .

Answer

The lengths of the horizontal and vertical legs of the triangle correspond to the -coordinate of the -intercept and the -coordinate of the -intercept. The area of a right triangle is half the product of the lengths of its legs and . The length of the vertical leg is , so, setting and , and solving for :

$\frac{1}{2} ab = A$

$\frac{1}{2} \cdot a \cdot 16 = 48$

$\frac{8a }{8}= \frac{48}{8}$

Therefore, the -intercept of the line containing the hypotenuse is . The slope of the line given the coordinates of its intercepts is

$m = - \frac{b}{a}$ .

substituting:

$m = - \frac{16}{6} = -\frac{8}{3}$ .

Substituting for and in the slope-intercept form of the equation of a line,

,

the line has equation

$y = -\frac{8}{3} x+16$ .

Substituting for and 6 for and solving for , we find the -coordinate

of the point on the line with -coordinate 6:

$6 = -\frac{8}{3} X+16$

$6 + \frac{8}{3} X - 6 = -\frac{8}{3} X+16 + \frac{8}{3} X - 6$

$\frac{8}{3} X = 10$

$\frac{3} {8}\cdot \frac{8}{3} X = \frac{3} {8}\cdot 10$

$X = \frac{15} {4} = 3 \frac{3}{4}$

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Question

Lines P and Q are parallel. Find the value of .

Answer

Since these are complementary angles, we can set up the following equation.

Now we will use the quadratic formula to solve for .

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-10\pm \sqrt{10^2-4(1)(-180)}}{2(1)}$

$x=\frac{-10\pm \sqrt{100+720}}{2}$

$x=\frac{-10\pm \sqrt{820}}{2}$

$x=\frac{-10\pm \sqrt{4\cdot 205}}{2}$

$x=\frac{-10\pm 2\sqrt{ 205}}{2}$

$x=-5\pm \sqrt{ 205}$

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Question

$\begin{tabular}{||c c||} \hline Time & Distance \ [0.5ex] \hline\hline 0&10 \ 1&4 \ 2&3 \ 2.5 & 1 \ 3&3\ 4&4 \ 5&10\ \hline \hline \end{tabular}$

The table and graph describe two different particle's travel over time. Which particle has a lower minimum?

Answer

This question is testing one's ability to compare the properties of functions when they are illustrated in different forms. This question specifically is asking for the examination and interpretation of two quadratic functions for which one is illustrated in a table format and the other is illustrated graphically.

Step 1: Identify the minimum of the table.

Using the table find the time value where the lowest distance exists.

$\begin{tabular}{||c c||} \hline Time & Distance \ [0.5ex] \hline\hline 0&10 \ 1&4 \ 2&3 \ {\color{Red} 2.5} & {\color{Red} 1} \ 3&3\ 4&4 \ 5&10\ \hline \hline \end{tabular}$

Recall that the time represents the values while the distance represents the values. Therefore the ordered pair for the minimum can be written as .

Step 2: Identify the minimum of the graph

Recall that the minimum of a cubic function is known as a local minimum. This occurs at the valley where the vertex lies.

For this particular graph the vertex is at .

Step 3: Compare the minimums from step 1 and step 2.

Compare the value coordinate from both minimums.

$-3<1\Rightarrow \textup{Graph}< \textup{Table}$

Therefore, the graph has the lowest minimum.

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