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

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.

Explanation

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.

Axes

Figure NOT drawn to scale.

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

Evaluate .

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

$X = 3 \frac{1}{2}$

$X = 3 \frac{1}{4}$

Explanation

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:

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

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

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

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

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

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

$\dpi{100} \small 3x-6=2y$

$\dpi{100} \small 8x=9y$

$\dpi{100} \small 8x+9=y$

$\dpi{100} \small 3x+6=2y$

$\dpi{100} \small 3x+6=y$

Explanation

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 $\dpi{100} \small 3x-6=2y$

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

Which point lies on this line?

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

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

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

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

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

Explanation

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

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

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

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

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

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

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

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

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

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

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

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

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

Which of these points is on that line?

(4,7)

(4,-7)

(-4,7)

(2,7)

(-2,7)

Explanation

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)

Trans

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

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

$x= \sqrt{ 205}$

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

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

$x=\pm \sqrt{ 205}$

Explanation

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

Now we will use the quadratic formula to solve for .

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

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

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

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

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

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

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

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 ?

-1

-2

Explanation

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.

At what point do these two lines intersect?

$(\frac{8}{15}, \frac{3}{15})$

$(-\frac{8}{15}, \frac{3}{15})$

None of the given answers

Explanation

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 $(\frac{8}{15}, \frac{3}{15})$

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

Explanation

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.

Solve the following system of equations:

–2x + 3y = 10

2x + 5y = 6

(2, 2)

(3, –2)

(–2, –2)

(3, 5)

(–2, 2)

Explanation

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)

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

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