SAT Math - How to find x or y intercept

Give the area of the triangle on the coordinate plane that is bounded by the axes and the line of the equation .

$56 \frac{1}{4}$

$112\frac{1}{2}$

Explanation

It is necessary to find the vertices of the triangle, each of which is a point at which two of the three lines intersect.

The two axes intersect at the origin, making this one vertex.

The other two points of intersection are the intercepts of the line of equation . Since this equation is in slope-intercept form , where is the -coordinate of the -intercept, then , and the -intercept is the point . The -coordinate of the -intercept, , can be found by setting and solving for :

$a = \frac{15}{2} = 7 \frac{1}{2}$ .

The three vertices are located at $(0,0), \left ( 7\frac{1}{2}, 0 \right ), (0. -15)$

The line in question is shown below, with the bounded triangle shaded in:

Triangle z

The lengths of its legs are equal to the absolute values of the nonzero coordinates of its two intercepts - $a' = 7 \frac{1}{2}, b' = 15$ . The area of this right triangle is half their product:

$A = \frac{1}{2} a 'b' = \frac{1}{2} \cdot 7\frac{1}{2} \cdot 15 = 56 \frac{1}{4}$ .

Determine the y-intercept of the following line:

$\dpi{100} \small 3x+6y=9$

$\dpi{100} \small 1.5$

$\dpi{100} \small 3$

$\dpi{100} \small 6$

$\dpi{100} \small 9$

$\dpi{100} \small \frac{1}{3}$

Explanation

The y-intercept occurs when $\dpi{100} \small x=0$

$\dpi{100} \small 3x+6y=9$

$\dpi{100} \small 3(0)+6y=9$

$\dpi{100} \small 0+6y=9$

$\dpi{100} \small y = \frac{9}{6}=1.5$

Give the area of the triangle on the coordinate plane that is bounded by the lines of the equations , and .

$60\frac{1}{6}$

$54 \frac{5}{8}$

$72 \frac{5}{6}$

Explanation

It is necessary to find the coordinates of the vertices of the triangle, each of which is the intersection of two of the three lines.

The intersection of the lines of the equations and can easily be found by noting that since , by substitution, , making the point of intersection .

The intersection of the lines of the equations and can be found by substituting for in the latter equation and evaluating :

The point of intersection is .

The intersection of the lines of the equations and can be found by substituting for in the latter equation and solving for :

$\frac{3x }{3}= \frac{7}{3}$

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

, so $y = 2\frac{1}{3}$ , and this point of intersection is $\left ( 2\frac{1}{3}, 2\frac{1}{3} \right )$ .

The lines in question are graphed below, and the triangle they bound is shaded:

Triangle z

We can take the vertical side as the base of the triangle; its length is the difference of the -coordinates:

The height is the horizontal distance from this side to the opposite side, which is the difference of the -coordinates:

$h= 2 \frac{1}{3} - (-4) = 6 \frac{1}{3}$

The area is half their product:

$A = \frac{1}{2} bh = \frac{1}{2} \cdot 19 \cdot 6 \frac{1}{3} =60\frac{1}{6}$

Find the x-intercepts of $25x^{2}+4y^{2} = 9$ .

$\pm \frac{3}{5}$

$\frac{3}{5}$

$5$

$2$

$\pm 5$

Explanation

To find the x-intercepts, plug $y=0$ into the equation and solve for $x$ .

$25x^{2} + 4\cdot 0^{2} = 9$

$25x^{2} = 9$

$x^{2} = \frac{9}{25}$

$x = \pm \frac{3}{5}$

Don't forget that there are two solutions, both negative and positive!

Where does the line given by $y=3(x-4)-9$ intercept the -axis?

$\frac{4}{3}$

Explanation

First, put in slope-intercept form.

$y=3x-21$ .

To find the -intercept, set and solve for .

Which of the following lines does not intersect the line ?

$y=\frac{1}{4}x-7$

$y=-\frac{1}{4}x+11$

Explanation

Parallel lines never intersect, so you are looking for a line that has the same slope as the one given. The slope of the given line is –4, and the slope of the line in y = –4_x_ + 5 is –4 as well. Since these two lines have equal slopes, they will run parallel and can never intersect.

If these three points are on a single line, what is the formula for the line?

(3,3)

(4,7)

(5,11)

y = 4x - 9

y = 4x + 31

y = 3x - 3

y = 3x - 9

y = 5x + 11