GMAT Math : Coordinate Geometry

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Example Questions

Example Question #1 : Calculating The Equation Of A Curve

If the -intercept and the slope are , what's the equation of the line in standard form?

Possible Answers:

x+y=0

x-y=0

x+y=1

x-y=1

x-y=-1

Correct answer:

x-y=1

Explanation:

Write the slope intercept formula.

y=mx+b

Convert the given x-intercept to a known point, which is (1,0) .

Substitute the given slope and the point to solve for the y-intercept.

0=(1)(1)+b

b=-1

Substitute the slope and y-intercept into the slope-intercept formula.

y=x-1

Add 1 on both sides of the equation, and subtract on both sides of the equation to find the equation in standard form.

x-y=1

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Example Question #2 : Calculating The Equation Of A Curve

Which of the following functions has as its graph a curve with (0, -4) , and (0,4) as its only two -intercepts?

Possible Answers:

$f(x) = x^{3} - 64$

$f(x) = x^{3}- 4 x^{2}-16 x - 64$

$f(x) = x^{3} + 64$

$f(x) = x^{3}- 4 x^{2}+16 x - 64$

$f(x) = x^{3}- 4 x^{2}-16 x + 64$

Correct answer:

$f(x) = x^{3}- 4 x^{2}-16 x + 64$

Explanation:

By the Fundamental Theorem of Algebra, a polynomial equation of degree 3 must have three solutions, or roots, but one root can be a double root or triple root. Since the polynomial here has two roots, and 4, one of these must be a double root. Since the leading term is $x^{2}$ , the equation must be

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

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

We rewrite both.

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

(x+4)(x+4)(x-4) = 0

$(x+4)(x^{2}-4^{2}) = 0$

$(x+4)(x^{2}-16) = 0$

$x \cdot x^{2}- x \cdot 16 + 4 \cdot x^{2}- 4 \cdot 16 = 0$

$x^{3}-16 x + 4 x^{2}- 64 = 0$

$x^{3}+ 4 x^{2}-16 x - 64 = 0$

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

(x+4)(x- 4)(x-4) = 0

$(x^{2}-16) (x-4)= 0$

$x \cdot x^{2}- 4 \cdot x^{2}- x \cdot 16 + 4 \cdot 16 = 0$

$x^{3}- 4 x^{2}-16 x + 64 = 0$

The correct response can be $f(x) = x^{3}+ 4 x^{2}-16 x - 64$ or $f(x) = x^{3}- 4 x^{2}-16 x + 64$ . The first is not among the choices, so the last is the correct choice.

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Example Question #5 : Calculating The Equation Of A Curve

Which of the following functions does not have as its graph a curve with (-5,0) as an -intercept?

Possible Answers:

$f(x) = x^{3}+1 5x^{2} + 75 x+125$

$f(x) = x^{3}-1 5x^{2} + 75 x-125$

$f(x) = x^{3}+ 5x^{2} -25 x-125$

$f(x) = x^{3}- 5x^{2} -25 x+125$

$f(x) = x^{3} +125$

Correct answer:

$f(x) = x^{3}-1 5x^{2} + 75 x-125$

Explanation:

We can evaluate f(-5) in each of the definitions of in the five choices. If f(-5) = 0 , (-5,0) is an -intercept.

$f(x) = x^{3}+ 5x^{2} -25 x-125$

$f(-5) = (-5)^{3}+ 5 \cdot (-5)^{2} -25 (-5)-125$

$= (-5)^{3}+ 5 \cdot 25 - (-125)-125$

= -125 +125 +125-125 = 0

$f(x) = x^{3}- 5x^{2} -25 x+125$

$f(-5) = (-5)^{3}- 5 \cdot (-5)^{2} -25 (-5)+125$

$= (-5)^{3}+ 5 \cdot 25 - (-125)-125$

= -125 +125 +125-125 = 0

$f(x) = x^{3} +125$

$f(-5) = (-5) ^{3} +125$

= -125+125 = 0

$f(x) = x^{3}+1 5x^{2} + 75 x+125$

$f(-5) = (-5)^{3}+1 5\cdot (-5)^{2} + 75 (-5)+125$

$f(-5) = -125+1 5\cdot 25 + (-375)+125$

f(-5) = -125+375-375+125 = 0

$f(x) = x^{3}-1 5x^{2} + 75 x-125$

$f(-5) = (-5)^{3}-1 5\cdot (-5)^{2} + 75 (-5)-125$

$f(-5) = -125-1 5\cdot 25 + (-375)-125$

f(-5) = -125-375-375-125 = -1,000

$f(x) = x^{3}-1 5x^{2} + 75 x-125$ does not have (-5,0) as an -intercept, so it is the correct choice.

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Example Question #6 : Calculating The Equation Of A Curve

A function is defined as

$f(x) = 2x^{4} + Bx^{3}+ Cx^{2}+ Dx +12$

where B,C,D are integer coefficients whose values (which might be positive, negative, or zero) are not given. Which of the following cannot be an -intercept of the graph of no matter what the values of those three coefficients are?

Possible Answers:

$\left ( \frac{1}{2}, 0 \right )$

(2,0)

$\left ( \frac{3}{2}, 0 \right )$

(3,0)

$\left ( \frac{1}{6}, 0 \right )$

Correct answer:

$\left ( \frac{1}{6}, 0 \right )$

Explanation:

Since the graph of a function has its -intercept at a point (a,0) if and only if f(a)= 0 , finding possible -intercepts of the graph of is equivalent to finding a solution of f(x)= 0 . Since has integer coefficients, then by the Rational Zeroes Theorem, any rational solutions to the equation

$f(x) = 2x^{4} + Bx^{3}+ Cx^{2}+ Dx +12$

must be the quotient, or the (negative) opposite of the quotient, of a factor of constant coefficient 12 - that is, an element of $\left \{ 1, 2, 3, 4, 6, 12 \right \}$ - and a factor of leading coefficient 2 - that is, an element of $\left \{ 1,2 \right \}$ . Since all of the choices are positive, we will only look at possible positive solutions.

The quotients of an element of the first set and an element of the last are:

$\frac{1}{1} = 1$ ; $\frac{2}{1} = 2$ ; $\frac{3}{1} = 3$ ; $\frac{4}{1} = 4$ ; $\frac{6}{1} = 6$ ; $\frac{12}{1} = 12$ ;

$\frac{1}{2}$ ; $\frac{2}{2} = 1$ ; $\frac{3}{2}$ ; $\frac{4}{2} = 2$ ; $\frac{6}{2} = 3$ ; $\frac{12}{2} = 6$

Eliminating duplicates, the set of possible positive rational solutions to f(x) = 0 is

$\left \{ \frac{1}{2},1, \frac{3}{2} , 2,3,4,6,12 \right \}$ .

Of the five choices, only $\frac{1}{6}$ does not appear in the set of possible rational solutions of f(x) = 0 , so of the five choices, only $\left ( \frac{1}{6}, 0 \right )$ cannot be an -intercept of the graph.

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Example Question #7 : Calculating The Equation Of A Curve

Between which two points is an -intercept of the graph of the function

$f(x) = x^{4}+3x^{2}-17x-115$

located?

Possible Answers:

Between (1,0) and (2,0)

Between (0,0) and (1,0)

Between (3,0) and (4,0)

Between (4,0) and (5,0)

Between (2,0) and (3,0)

Correct answer:

Between (3,0) and (4,0)

Explanation:

As a polynomial function, has a continuous graph. By the Intermediate Value Theorem, if f(a) and f(b) are of different sign, then f(c) = 0 for some $c \in (a,b)$ - that is, the graph of has an -intercept between (a,0) and (b,0) . Evaluate for all x= 0,1,2,3,4,5 and observe between which two integers the sign changes.

$f(x) = x^{4}+3x^{2}-17x-115$

$f(0) = 0^{4}+3 \cdot 0^{2}-17\cdot 0-115$

= 0+0-0-115

= -115

$f(1) =1^{4}+3\cdot 1^{2}-17 \cdot 1-115$

= 1 +3 -17-115

= -128

$f(2) = 2^{4}+3\cdot 2^{2}-17\cdot 2-115$

$= 16+3\cdot 4 -34 -115$

= 16+12 -34 -115

= -121

$f(3) =3 ^{4}+3 \cdot 3^{2}-17 \cdot 3-115$

$= 81 + 3 \cdot 9 - 51 -115$

= 81 + 27 -51 -115

= -58

$f(4) = 4 ^{4}+3\cdot 4^{2}-17\cdot 4 -115$

$= 256 +3\cdot 16 -68-115$

= 256 +48 -68-115

121

$f(5) = 5^{4}+3 \cdot 5^{2}-17 \cdot 5 -115$

$= 625 +3 \cdot 25 -85 -115$

= 625 +75 -85 -115

=500

Since f(3)< 0 and f(4) > 0 , the -intercept is between (3,0) and (4,0) .

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Example Question #8 : Calculating The Equation Of A Curve

Only one of the following equations has a graph with an -intercept between $\left ( \frac{1}{2},0 \right )$ and (1 ,0) . Which one?

Possible Answers:

$f(x) = x^{4}-x^{3}+4x-3$

$f(x) = x^{4}-x^{3}+2x-3$

$f(x) = x^{4}-x^{3}+16x-3$

$f(x) = x^{4}-x^{3}+8x-3$

$f(x) = x^{4}-x^{3}+ x-3$

Correct answer:

$f(x) = x^{4}-x^{3}+4x-3$

Explanation:

The Intermediate Value Theorem states that if is a continuous function, as all five of the polynomial functions in the given choices are, and $f\left ( \frac{1}{2} \right )$ and f(1) are of different sign, then the graph of has an -intercept on the interval $\left ( \frac{1}{2}, 1 \right )$ .

We evaluate $f\left ( \frac{1}{2} \right )$ and f(1) for each of the five choices to find the one for which the two have different sign.

$f(x) = x^{4}-x^{3}+ x-3$

$f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+ \frac{1}{2} -3 = \frac{1}{16} - \frac{1}{8}+ \frac{1}{2}-3=-2\frac{9}{16}$

$f(1) = 1^{4}-1^{3}+ 1-3 = 1-1+1-3 = -2$

$f\left ( \frac{1}{2} \right )$ and f(1) are both negative.

$f(x) = x^{4}-x^{3}+ 2 x-3$

$f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+2 \left (\frac{1}{2} \right )-3 = \frac{1}{16} - \frac{1}{8}+1-3=-2\frac{1}{16}$

$f(1) = 1^{4}-1^{3}+ 2 \cdot 1-3 = 1-1+2-3 = -1$

$f\left ( \frac{1}{2} \right )$ and f(1) are both negative.

$f(x) = x^{4}-x^{3}+ 4 x-3$

$f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+4 \left (\frac{1}{2} \right )-3 = \frac{1}{16} - \frac{1}{8}+2-3=-1\frac{1}{16}$

$f(1) = 1^{4}-1^{3}+ 4 \cdot 1-3 = 1-1+4-3 = 1$

$f\left ( \frac{1}{2} \right )$ and f(1) are of different sign.

$f(x) = x^{4}-x^{3}+ 8 x-3$

$f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+8\left (\frac{1}{2} \right )-3 = \frac{1}{16} - \frac{1}{8}+4-3= \frac{15}{16}$

$f(1) = 1^{4}-1^{3}+ 8 \cdot 1-3 = 1-1+8-3 = 5$

$f\left ( \frac{1}{2} \right )$ and f(1) are both positive.

$f(x) = x^{4}-x^{3}+ 16x-3$

$f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+16\left (\frac{1}{2} \right )-3 = \frac{1}{16} - \frac{1}{8}+8-3=4 \frac{15}{16}$

$f(1) = 1^{4}-1^{3}+ 16 \cdot 1-3 = 1-1+16-3 = 13$

$f\left ( \frac{1}{2} \right )$ and f(1) are both positive.

$f(x) = x^{4}-x^{3}+ 4 x-3$ is the function in which $f\left ( \frac{1}{2} \right )$ and f(1) are of different sign, so it is represented by a graph with an -intercept between $\left ( \frac{1}{2},0 \right )$ and (1 ,0) . This is the correct choice.

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Example Question #9 : Calculating The Equation Of A Curve

Which of the following functions has as its graph a curve with -intercepts (-2, 0) , (2,0) , and (3,0) ?

Possible Answers:

$f(x) = x^{3}+x^{2}-8x-12$

$f(x) = x^{3}-3x^{2}-4x+12$

$f(x) = x^{3}+3x^{2}-4x-12$

$f(x) = x^{3}- x^{2}-8x+12$

$f(x) = x^{3}-7x^{2}+16x-12$

Correct answer:

$f(x) = x^{3}-3x^{2}-4x+12$

Explanation:

A polynomial equation of degree 3 with solution set $\left \{ -2, 2, 3\right \}$ and leading term $x^{2}$ takes the form

(x- (-2)) (x-2)(x-3)= 0

We can rewrite this as follows:

(x+2) (x-2)(x-3)= 0

$(x^{2}-2^{2})(x-3)= 0$

$(x^{2}-4)(x-3)= 0$

$x^{2}\cdot x- x^{2} \cdot 3 - 4\cdot x + 4 \cdot 3 = 0$

$x^{3} - 3 x^{2} - 4 x + 12 = 0$

The correct response is $f(x)= x^{3} - 3 x^{2} - 4 x + 12$ .

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Example Question #721 : Geometry

One of the diameters of a circle has endpoints (4, 5) and (10, 1). What is the equation of this circle?

Possible Answers:

$\small \small \small (x - 3)^{2} + (y - 7)^{2} = 13$

$\small \small (x - 7)^{2} + (y - 3)^{2} = 169$

$\small \small \small \small (x - 3)^{2} + (y - 7)^{2} = 169$

$\small 3x^{2} + 9y^{2} = 169$

$\small (x - 7)^{2} + (y - 3)^{2} = 13$

Correct answer:

$\small (x - 7)^{2} + (y - 3)^{2} = 13$

Explanation:

The equation of a circle with center (h,k) and radius is

$\small \small \small \small (x - h)^{2} + (y - k)^{2} = r^{2}$

The center is the midpoint of any diameter, so to find the center, we use the midpoint formula:

$\small \small h= \small \frac{x_{1}+x_{2}}{2}= \small \frac{4+10}{2} = 7$

$\small \small \small k= \small \frac{y_{1}+y_{2}}{2}= \small \frac{5+1}{2} = 3$

The center is (7,3). The radius is the distance between (7,3) and (10,1), so we use the distance formula:

$\small r = \sqrt{(10-7)^{2}+(1-3))^{2}} = \sqrt{3^{2}+(-2)^{2}} = \sqrt{13}$

So $\small r^{2} = 13$ , and the equation of the circle is

$\small (x - 7)^{2} + (y - 3)^{2} = 13$

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Example Question #1 : Calculating The Equation Of A Circle

In the coordinate plane, a circle has center $\left ( 5,2 \right )$ and passes through the point $\left ( -1,-6 \right )$ . What is the area of the circle?

Possible Answers:

$100\pi$

$36\pi$

$80\pi$

$4\sqrt{2}\pi$

$10\pi$

Correct answer:

$100\pi$

Explanation:

The distance of the two points is $\sqrt{(5-(-1))^2+(2-(-6))^2}=10$ .

So 10 is the radius of the circle. Then we can calculate the area:

$A=\pi r^{2}=\pi \cdot 10^{2}=100\pi$ .

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Example Question #731 : Geometry

A circle on the coordinate plane has area $64\pi$ ; its center is the origin. Which of the following is the equation of this circle?

Possible Answers:

$x ^{2} + y ^{2} = 64$

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

$x ^{2} + y ^{2} = 4,096$

$x ^{2} + y ^{2} = 256$

$x ^{2} + y ^{2} = 16$

Correct answer:

$x ^{2} + y ^{2} = 64$

Explanation:

The equation of a circle with center at the origin is

$x ^{2} + y ^{2} = r^{2}$

where is the radius of the circle. The area of the circle is $A = \pi r ^{2}$ .

Since the area of the circle in the question is $64\pi$ , we can solve for $r ^{2}$ :

$A = \pi r ^{2} = 64 \pi$

$\pi r ^{2} \div \pi = 64 \pi \div \pi$

$r ^{2} = 64$

The equation is

$x ^{2} + y ^{2} = 64$

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GMAT Math : Coordinate Geometry

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1 : Calculating The Equation Of A Curve

Example Question #2 : Calculating The Equation Of A Curve

Example Question #5 : Calculating The Equation Of A Curve

Example Question #6 : Calculating The Equation Of A Curve

Example Question #7 : Calculating The Equation Of A Curve

Example Question #8 : Calculating The Equation Of A Curve

Example Question #9 : Calculating The Equation Of A Curve

Example Question #721 : Geometry

Example Question #1 : Calculating The Equation Of A Circle

Example Question #731 : Geometry

All GMAT Math Resources

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