DSQ: Calculating the equation of a line - GMAT Quantitative

Card 0 of 35

Question

Find the equation of linear function given the following statements.

I) $d(g)\perp h(j)=5x+7$

II) intercepts the x-axis at 9.

Answer

To find the equation of a linear function, we need some combination of slope and a point.

Statement I gives us a clue to find the slope of the desired function. It must be the opposite reciprocal of the slope of . This makes the slope of equal to $-\frac{1}{5}$

Statement II gives us a point on our desired function, .

Using slope-intercept form, we get the following:

$0=-\frac{1}{5}\cdot 6+b$

$b=\frac{6}{5}$

So our equation is as follows

$y=-\frac{1}{5}x+\frac{6}5$

Compare your answer with the correct one above

Question

Find the equation for linear function .

I) $f(x)\parallel p(x)$ and

II) $p(\frac{1}{2})=-4$

Answer

Find the equation for linear function p(x)

I) $f(x)\parallel p(x)$ and

II) $p(\frac{1}{2})=-4$

To begin:

I) Tells us that p(x) must have a slope of 16

II) Tells us a point on p(x). Plug it in and solve for b:

$-4=\frac{1}{2}*16+b$

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Question

There are two lines in the xy-coordinate plane, a and b, both with positive slopes. Is the slope of a greater than the slope of b?

1)The square of the x-intercept of a is greater than the square of the x-intercept of b.

Lines a and b have an intersection at

Answer

Given that the square of a negative is still positive, it is possible for a to have an x-intercept that is negative, while still having a positive slope. The example above shows how the square of the x-intercept for line a could be greater, while having still giving line a a slope that is less than that of b.

Compare your answer with the correct one above

Question

Line j passes through the point . What is the equation of line j?

Line j is perpindicular to the line defined by $y = -\frac{1}{2}x+3$

Line j has an x-intercept of

Answer

Either statement is sufficient.

Line j, as a line, has an equation of the form

Statement 1 gives the equation of a perpindicular line, so the slopes of the two lines are negative reciprocals of each other:

Statement 2 allows the slope to be found using rise over run:

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{6-0}{4-1}=3$

Then, since the x-intercept is known:

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Question

Give the equation of a line.

Statement 1: The line interects the graph of the equation $y = x^{2}- 7x + 6$ on the -axis.

Statement 2: The line interects the graph of the equation $y = x^{2}- 7x + 6$ on the -axis.

Answer

Assume both statements to be true. Then the line shares its - and -intercepts with the graph of $y = x^{2}- 7x + 6$ , which is a parabola. The common -intercept can be found by setting and solving for :

$y = x^{2}- 7x + 6$

$y = 0^{2}- 7\cdot 0 + 6 = 0 - 0 + 6 = 6$ ,

making the -intercept of the parabola, and that of the line, .

The common -intercept can be found by setting and solving for :

$0= x^{2}- 7x + 6$

, in which case , or

, in which case ,

The parabola therefore has two -intercepts, and , so it is not clear which one is the -intercept of the line. Therefore, the equation of the line is also unclear.

Compare your answer with the correct one above

Question

Find the equation of linear function given the following statements.

I) $d(g)\perp h(j)=5x+7$

II) intercepts the x-axis at 9.

Answer

To find the equation of a linear function, we need some combination of slope and a point.

Statement I gives us a clue to find the slope of the desired function. It must be the opposite reciprocal of the slope of . This makes the slope of equal to $-\frac{1}{5}$

Statement II gives us a point on our desired function, .

Using slope-intercept form, we get the following:

$0=-\frac{1}{5}\cdot 6+b$

$b=\frac{6}{5}$

So our equation is as follows

$y=-\frac{1}{5}x+\frac{6}5$

Compare your answer with the correct one above

Question

Find the equation for linear function .

I) $f(x)\parallel p(x)$ and

II) $p(\frac{1}{2})=-4$

Answer

Find the equation for linear function p(x)

I) $f(x)\parallel p(x)$ and

II) $p(\frac{1}{2})=-4$

To begin:

I) Tells us that p(x) must have a slope of 16

II) Tells us a point on p(x). Plug it in and solve for b:

$-4=\frac{1}{2}*16+b$

Compare your answer with the correct one above

Question

There are two lines in the xy-coordinate plane, a and b, both with positive slopes. Is the slope of a greater than the slope of b?

1)The square of the x-intercept of a is greater than the square of the x-intercept of b.

Lines a and b have an intersection at

Answer

Given that the square of a negative is still positive, it is possible for a to have an x-intercept that is negative, while still having a positive slope. The example above shows how the square of the x-intercept for line a could be greater, while having still giving line a a slope that is less than that of b.

Compare your answer with the correct one above

Question

Line j passes through the point . What is the equation of line j?

Line j is perpindicular to the line defined by $y = -\frac{1}{2}x+3$

Line j has an x-intercept of

Answer

Either statement is sufficient.

Line j, as a line, has an equation of the form

Statement 1 gives the equation of a perpindicular line, so the slopes of the two lines are negative reciprocals of each other:

Statement 2 allows the slope to be found using rise over run:

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{6-0}{4-1}=3$

Then, since the x-intercept is known:

Compare your answer with the correct one above

Question

Give the equation of a line.

Statement 1: The line interects the graph of the equation $y = x^{2}- 7x + 6$ on the -axis.

Statement 2: The line interects the graph of the equation $y = x^{2}- 7x + 6$ on the -axis.

Answer

Assume both statements to be true. Then the line shares its - and -intercepts with the graph of $y = x^{2}- 7x + 6$ , which is a parabola. The common -intercept can be found by setting and solving for :

$y = x^{2}- 7x + 6$

$y = 0^{2}- 7\cdot 0 + 6 = 0 - 0 + 6 = 6$ ,

making the -intercept of the parabola, and that of the line, .

The common -intercept can be found by setting and solving for :

$0= x^{2}- 7x + 6$

, in which case , or

, in which case ,

The parabola therefore has two -intercepts, and , so it is not clear which one is the -intercept of the line. Therefore, the equation of the line is also unclear.

Compare your answer with the correct one above

Question

Find the equation of linear function given the following statements.

I) $d(g)\perp h(j)=5x+7$

II) intercepts the x-axis at 9.

Answer

To find the equation of a linear function, we need some combination of slope and a point.

Statement I gives us a clue to find the slope of the desired function. It must be the opposite reciprocal of the slope of . This makes the slope of equal to $-\frac{1}{5}$

Statement II gives us a point on our desired function, .

Using slope-intercept form, we get the following:

$0=-\frac{1}{5}\cdot 6+b$

$b=\frac{6}{5}$

So our equation is as follows

$y=-\frac{1}{5}x+\frac{6}5$

Compare your answer with the correct one above

Question

Find the equation for linear function .

I) $f(x)\parallel p(x)$ and

II) $p(\frac{1}{2})=-4$

Answer

Find the equation for linear function p(x)

I) $f(x)\parallel p(x)$ and

II) $p(\frac{1}{2})=-4$

To begin:

I) Tells us that p(x) must have a slope of 16

II) Tells us a point on p(x). Plug it in and solve for b:

$-4=\frac{1}{2}*16+b$

Compare your answer with the correct one above

Question

There are two lines in the xy-coordinate plane, a and b, both with positive slopes. Is the slope of a greater than the slope of b?

1)The square of the x-intercept of a is greater than the square of the x-intercept of b.

Lines a and b have an intersection at

Answer

Given that the square of a negative is still positive, it is possible for a to have an x-intercept that is negative, while still having a positive slope. The example above shows how the square of the x-intercept for line a could be greater, while having still giving line a a slope that is less than that of b.

Compare your answer with the correct one above

Question

Line j passes through the point . What is the equation of line j?

Line j is perpindicular to the line defined by $y = -\frac{1}{2}x+3$

Line j has an x-intercept of

Answer

Either statement is sufficient.

Line j, as a line, has an equation of the form

Statement 1 gives the equation of a perpindicular line, so the slopes of the two lines are negative reciprocals of each other:

Statement 2 allows the slope to be found using rise over run:

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{6-0}{4-1}=3$

Then, since the x-intercept is known:

Compare your answer with the correct one above

Question

Give the equation of a line.

Statement 1: The line interects the graph of the equation $y = x^{2}- 7x + 6$ on the -axis.

Statement 2: The line interects the graph of the equation $y = x^{2}- 7x + 6$ on the -axis.

Answer

Assume both statements to be true. Then the line shares its - and -intercepts with the graph of $y = x^{2}- 7x + 6$ , which is a parabola. The common -intercept can be found by setting and solving for :

$y = x^{2}- 7x + 6$

$y = 0^{2}- 7\cdot 0 + 6 = 0 - 0 + 6 = 6$ ,

making the -intercept of the parabola, and that of the line, .

The common -intercept can be found by setting and solving for :

$0= x^{2}- 7x + 6$

, in which case , or

, in which case ,

The parabola therefore has two -intercepts, and , so it is not clear which one is the -intercept of the line. Therefore, the equation of the line is also unclear.

Compare your answer with the correct one above

Question

Find the equation of linear function given the following statements.

I) $d(g)\perp h(j)=5x+7$

II) intercepts the x-axis at 9.

Answer

To find the equation of a linear function, we need some combination of slope and a point.

Statement I gives us a clue to find the slope of the desired function. It must be the opposite reciprocal of the slope of . This makes the slope of equal to $-\frac{1}{5}$

Statement II gives us a point on our desired function, .

Using slope-intercept form, we get the following:

$0=-\frac{1}{5}\cdot 6+b$

$b=\frac{6}{5}$

So our equation is as follows

$y=-\frac{1}{5}x+\frac{6}5$

Compare your answer with the correct one above

Question

Find the equation for linear function .

I) $f(x)\parallel p(x)$ and

II) $p(\frac{1}{2})=-4$

Answer

Find the equation for linear function p(x)

I) $f(x)\parallel p(x)$ and

II) $p(\frac{1}{2})=-4$

To begin:

I) Tells us that p(x) must have a slope of 16

II) Tells us a point on p(x). Plug it in and solve for b:

$-4=\frac{1}{2}*16+b$

Compare your answer with the correct one above

Question

There are two lines in the xy-coordinate plane, a and b, both with positive slopes. Is the slope of a greater than the slope of b?

1)The square of the x-intercept of a is greater than the square of the x-intercept of b.

Lines a and b have an intersection at

Answer

Given that the square of a negative is still positive, it is possible for a to have an x-intercept that is negative, while still having a positive slope. The example above shows how the square of the x-intercept for line a could be greater, while having still giving line a a slope that is less than that of b.

Compare your answer with the correct one above

Question

Line j passes through the point . What is the equation of line j?

Line j is perpindicular to the line defined by $y = -\frac{1}{2}x+3$

Line j has an x-intercept of

Answer

Either statement is sufficient.

Line j, as a line, has an equation of the form

Statement 1 gives the equation of a perpindicular line, so the slopes of the two lines are negative reciprocals of each other:

Statement 2 allows the slope to be found using rise over run:

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{6-0}{4-1}=3$

Then, since the x-intercept is known:

Compare your answer with the correct one above

Question

Give the equation of a line.

Statement 1: The line interects the graph of the equation $y = x^{2}- 7x + 6$ on the -axis.

Statement 2: The line interects the graph of the equation $y = x^{2}- 7x + 6$ on the -axis.

Answer

Assume both statements to be true. Then the line shares its - and -intercepts with the graph of $y = x^{2}- 7x + 6$ , which is a parabola. The common -intercept can be found by setting and solving for :

$y = x^{2}- 7x + 6$

$y = 0^{2}- 7\cdot 0 + 6 = 0 - 0 + 6 = 6$ ,

making the -intercept of the parabola, and that of the line, .

The common -intercept can be found by setting and solving for :

$0= x^{2}- 7x + 6$

, in which case , or

, in which case ,

The parabola therefore has two -intercepts, and , so it is not clear which one is the -intercept of the line. Therefore, the equation of the line is also unclear.

Compare your answer with the correct one above

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