GMAT Math : Geometry

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Example Questions

Example Question #161 : Coordinate Geometry

The vertices of a triangle on the coordinate plane are the vertices and the -intercepts of the graph of the equation

$y = x^{2} - 10x + 24$ .

What is the area of this triangle?

Possible Answers:

$2 \frac{1}{2}$

$1 2 \frac{1}{2}$

Correct answer:

Explanation:

The vertex of the graph of a quadratic equation $y = a x^{2} + bx + c$ can be found by setting $x = -\frac{b}{2a}$ , substituting, and solving for . Setting a = 1, b = -10 , this is

$x = -\frac{b}{2a}$

$x = -\frac{-10}{2 \cdot 1} = 5$

Substitute:

$y = x^{2} - 10x + 24$

$y = 5 ^{2} - 10 \cdot 5 + 24 = 25 - 50 + 24 = -1$

The vertex is the point (5, -1) .

The -intercepts can be found by y= 0 and solving for :

$y = x^{2} - 10x + 24 = 0$

(x-4)(x-6) = 0

x- 4= 0 , in which case x = 4 , or

x-6 = 0 , in which case x = 6 .

The -intercepts are (4, 0) and (6, 0) .

The triangle with vertices (5, -1) , (4, 0) , and (6, 0) is below:

Triangle 3

If we take the base to be the horizontal segment connecting the -intercepts, then

b = 6-4 = 2 .

The height is therefore the vertical distance from this segment to the vertex, which is

h = |-1| = 1

The area of the triangle is half their product:

$A = \frac{1}{2} bh = \frac{1}{2} \cdot 2 \cdot 1 = 1$ .

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Example Question #1008 : Problem Solving Questions

A triangle on the coordinate plane has as its vertices the -intercept and -intercepts of the graph of the equation

$y = x^{2} -6x-16$ .

What is the area of this triangle?

Possible Answers:

250

125

160

Correct answer:

Explanation:

The -intercept of the graph of $y = x^{2} +6x-16$ , a parabola, can be found by setting x = 0 and solving for :

$y = x^{2} -6x-16$

$y =0 ^{2} -6 \cdot 0 -16$

y =0- 0 -16

y = -16

The -intercept is (0, -16) .

The -intercepts can be found by y= 0 and solving for :

$y = x^{2} -6x-16$

$x^{2} -6x-16 = 0$

(x-8)(x+2) = 0

x- 8 = 0 , in which case x = 8 , or

x+2 = 0 , in which case x = -2 .

The two -intercepts are (-2, 0) and (8, 0) .

The triangle is shown below:

Triangle 3

If the segment connecting the points on the -axes is taken as the base, then

b = 8 - (-2) = 10

The height is therefore the vertical distance from this segment to the point on the -axis, which is

h = |16| = 16

The area is half their product, or

$A = \frac{1}{2} bh = \frac{1}{2} \cdot 10 \cdot 16 = 80$ .

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Example Question #61 : Coordinate Geometry

$f(x)= Ax^{2}+ Bx+ C$ has as its graph a vertical parabola on the coordinate plane. You are given that A= 6 , but you are given neither nor .

Which of the following can you determine without knowing the values of and ?

I) Whether the curve opens upward or opens downward

II) The location of the vertex

III) The location of the -intercept

IV) The locations of the -intercepts, if there are any

V) The equation of the line of symmetry

Possible Answers:

II and V only

III only

I only

V only

III and IV only

Correct answer:

I only

Explanation:

I) The orientation of the parabola is determined solely by the value of . Since A= 6 > 0 , the parabola can be determined to open upward.

II and V) The -coordinate of the vertex is $-\frac{B}{2A}$ ; since you are not given , you cannot find this. Also, since the line of symmetry has equation $x=-\frac{B}{2A}$ , for the same reason, you cannot find this either.

III) The -intercept is the point at which x = 0 ; by substitution, it can be found to be at (0,C) . is unknown, so the -intercept cannot be found.

IV) The -intercept(s), if any, are the point(s) at which $f(x)= Ax^{2}+ Bx+ C = 0$ . This is solvable using the quadratic formula

$x= \frac{-B \pm \sqrt{B^{2}-4AC}}{2A}$

Since all three of A,B, and must be known for this to be evaluated, and only is known, the -intercept(s) cannot be identified.

The correct response is I only.

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Example Question #771 : Geometry

The parabolas of the functions f(x) and g(x) on the coordinate plane have the same vertex.

If we define $f(x) = \frac{4}{5}(x-9)^{2}+ 6$ , which of the following is a possible equation for g(x) ?

Screen shot 2016 02 10 at 12.25.12 pm

Possible Answers:

$g(x) = \frac{4}{5}(x-7)^{2}+ 6$

$g(x) = \frac{4}{5}(x-9)^{2}-2$

$g(x) = \frac{8}{5}(x-18)^{2}+ 12$

None of the other responses gives a correct answer.

$g(x) = \frac{3}{5}(x-9)^{2}+ 6$

Correct answer:

$g(x) = \frac{3}{5}(x-9)^{2}+ 6$

Explanation:

The eqiatopm of f(x) is given in the vertex form

$f(x) =a(x-h)^{2}+ k$ ,

so the vertex of its parabola is (h,k ) . The graphs of f(x) and g(x) are parabolas with the same vertex, so they must have the same values for and .

For the function $f(x) = \frac{4}{5}(x-9)^{2}+ 6$ , h = 9 and k = 6 .

Screen shot 2016 02 10 at 12.25.12 pm

Of the five choices, the only equation of g(x) that has these same values, and that therefore has a parabola with the same vertex, is $g(x) = \frac{3}{5}(x-9)^{2}+ 6$ .

Screen shot 2016 02 10 at 12.27.19 pm

To verify, graph both functions on the same grid.

Screen shot 2016 02 10 at 12.28.14 pm

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Example Question #1 : Graphing A Line

A line has slope $\frac{3}{4}$ . Which of the following could be its - and $y\,$ -intercepts, respectively?

Possible Answers:

$\left (12,0 \right )$ and $\left ( 0, 9 \right )$

$\left ( -8, 0 \right )$ and $\left ( 0, -6\right )$

None of the other responses gives a correct answer.

$\left (- 9, 0 \right )$ and $\left ( 0,-12 \right )$

$\left (6, 0 \right )$ and $\left ( 0, 8\right )$

Correct answer:

None of the other responses gives a correct answer.

Explanation:

Let (a,0) and (0,b) be the - and $y\,$ -intercepts, respectively, of the line. Then the slope of the line is $-\frac{b}{a}$ , or, equilvalently, $- (b\div a)$ .

We do not need to find the actual slopes of the four choices if we observe that in each case, and are of the same sign. Since the quotient of two numbers of the same sign is positive, it follows that $- (b\div a)$ is negative, and therefore, none of the pairs of intercepts can be those of a line with positive slope $\frac{3}{4}$ .

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Example Question #2 : Graphing A Line

A line has slope $-\frac{4}{3}$ . Which of the following could be its - and $y\,$ -intercepts, respectively?

Possible Answers:

$\left ( \frac{6}{7},0 \right )$ and $\left ( 0, \frac{4}{7} \right )$

$\left ( \frac{3}{7},0 \right )$ and $\left ( 0, \frac{4}{7} \right )$

$\left ( \frac{5}{6}, 0 \right )$ and $\left ( 0, \frac{5}{8}\right )$ :

None of the other responses gives a correct answer.

$\left ( \frac{4}{5}, 0 \right )$ and $\left ( 0, \frac{3}{5}\right )$

Correct answer:

$\left ( \frac{3}{7},0 \right )$ and $\left ( 0, \frac{4}{7} \right )$

Explanation:

Let (a,0) and (0,b) be the - and $y\,$ -intercepts, respectively, of the line. Then the slope of the line is $-\frac{b}{a}$ , or, equilvalently, $- (b\div a)$ .

We can examine the intercepts in each choice to determine which set meets these conditions.

$\left ( \frac{5}{6}, 0 \right )$ and $\left ( 0, \frac{5}{8}\right )$ :

Slope: $- \left (b\div a \right )=- \left (\frac{5}{8}\div \frac{5}{6} \right )= - \frac{3}{4}$

$\left ( \frac{6}{7},0 \right )$ and $\left ( 0, \frac{4}{7} \right )$

Slope: $- \left (b\div a \right )=- \left ( \frac{4}{7}\div \frac{6}{7} \right )=-\frac{2}{3}$

$\left ( \frac{3}{7},0 \right )$ and $\left ( 0, \frac{4}{7} \right )$

Slope: $- \left (b\div a \right )= - \left ( \frac{4}{7}\div \frac{3}{7} \right )= -\frac{4}{3}$

$\left ( \frac{4}{5}, 0 \right )$ and $\left ( 0, \frac{3}{5}\right )$

Slope: $- \left (b\div a \right )=- \left ( \frac{3}{5}\div \frac{4}{5} \right )= -\frac{3}{4}$

$\left ( \frac{3}{7},0 \right )$ and $\left ( 0, \frac{4}{7} \right )$ comprise the correct choice.

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Example Question #3 : Graphing A Line

Line_1

Which of the following equations can be graphed with a line perpendicular to the green line in the above figure, and with the same -intercept?

Possible Answers:

3x+5y = 9

-5 x +3y = 15

5x-3y = 15

3 x - 5y = 9

-3 x +5y = 25

Correct answer:

3 x - 5y = 9

Explanation:

The slope of the green line can be calculated by noting that the - and $y\,$ -intercepts of the line are, respectively, (3,0) and (0,5) . If (a,0) and (0,b) be the - and $y\,$ -intercepts, respectively, of a line, the slope of the line is $-\frac{b}{a}$ . This makes the slope of the green line $-\frac{5}{3}$ .

Any line perpendicular to this line must have as its slope the opposite reciprocal of this, or $\frac{3}{5}$ . Since the desired line must also have -intercept (3,0) , the equation of the line, in point=slope form, is

$y - y_{1} = m (x-x_{1})$

$y -0 = \frac{3}{5} (x-3)$

which can be simplified as

$y = \frac{3}{5} x-\frac{9}{5}$

$5 \cdot y =5 \cdot \left ( \frac{3}{5} x-\frac{9}{5} \right )$

5 y =3 x-9

5 y- 5y + 9 =3 x-9 - 5y + 9

3 x - 5y = 9

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Example Question #4 : Graphing A Line

A line passes through the vertex and the $y\:$ -intercept of the parabola of the equation $y = x^{2} - 8x+ 15$ . What is the equation of the line?

Possible Answers:

y = -x+3

y = x+5

y = -6x+4

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

y = -4x+15

Correct answer:

y = -4x+15

Explanation:

To locate the $y \:$ -intercept of the equation $y = x^{2} - 8x+ 15$ , substitute 0 for :

$y = x^{2} - 8x+ 15$

$y = 0^{2} - 8 \cdot 0+ 15$

y = 15

The $y \:$ -intercept of the parabola is (0,15) .

The vertex of the parabola of an equation of the form $y = ax^{2}+ bx + c$ has -coordinate $-\frac{b}{2a}$ . Here, we substitute a = 1, b = -8 , to obtain -coordinate

$-\frac{b}{2a} = -\frac{-8}{2 (1)} = 4$ .

To find the $y \:$ -coordinate, substitute this for :

$y = x^{2} - 8x+ 15$

$y = 4^{2} - 8 \cdot 4+ 15 = 16 - 32 +15 = -1$

The vertex is (4,-1) .

The line includes points (0,15) and (4,-1) ; apply the slope formula:

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

$m = \frac{15-(-1)}{0-4} = \frac{16}{-4} = -4$

The slope is , and the $y \;$ -intercept is (0,15) ; in the slope-intercept form y = mx+b , substitute for m = -4 and b = 15 . The equation of the line is y = -4x+15 .

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Example Question #5 : Graphing A Line

Give the equation of a line with undefined slope that passes through the vertex of the graph of the equation $y = - x^{2}- 14 x +16$ .

Possible Answers:

x = 16

y = 65

y = -7

x =- 7

y = 16

Correct answer:

x =- 7

Explanation:

A line with undefined slope is a vertical line, and its equation is x = a for some , so the -coordinate of all points it passes through is . If it goes through the vertex of a parabola (h,k ) , then the line has the equation x=h . Therefore, all we need to find is the -coordinate of the vertex of the parabola.

The vertex of the parabola of the equation $y = ax^{2}+ bx+c$ has as its -coordinate $h=-\frac{b}{2a}$ , which, for the parabola of the equation $y = - x^{2}- 14 x +16$ , can be found by setting a= -1, b=-14 :

$h = -\frac{b}{2a} = -\frac{-14}{2(-1)} = -7$

The desired line is x =- 7 .

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Example Question #2 : Graphing A Line

A line has slope 4. Which of the following could be its - and $y\,$ -intercepts, respectively?

Possible Answers:

(2.7, 0) and (0, -10.8)

(4.3, 0) and (0, 17.2)

None of the other responses gives a correct answer.

(12.4, 0) and (0, -3.1)

(12.8 , 0) and (0, 3.2)

Correct answer:

(2.7, 0) and (0, -10.8)

Explanation:

Let (a,0) and (0,b) be the - and $y\,$ -intercepts, respectively, of the line. Then the slope of the line is $-\frac{b}{a}$ , or, equilvalently, $- (b\div a)$ .

We can examine the intercepts in each choice to determine which set meets these conditions.

(12.8 , 0) and (0, 3.2)

Slope: $- \left (b\div a \right )=- \left (3.2\div 12.8 \right )=-0.25$

(4.3, 0) and (0, 17.2)

Slope: $- \left (b\div a \right )=- \left (17.2\div 4.3 \right )=-4$

(12.4, 0) and (0, -3.1)

Slope: $- \left (b\div a \right )= - \left (-3.1 \div 12.4 \right )= 0.25$

(2.7, 0) and (0, -10.8)

Slope: $- \left (b\div a \right )=- \left (-10.8\div 2.7 \right )=4$

(2.7, 0) and (0, -10.8) comprise the correct choice, since a line passing through these points has the correct slope.

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GMAT Math : Geometry

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #161 : Coordinate Geometry

Example Question #1008 : Problem Solving Questions

Example Question #61 : Coordinate Geometry

Example Question #771 : Geometry

Example Question #1 : Graphing A Line

Example Question #2 : Graphing A Line

Example Question #3 : Graphing A Line

Example Question #4 : Graphing A Line

Example Question #5 : Graphing A Line

Example Question #2 : Graphing A Line

All GMAT Math Resources

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