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# Line A has a slope of -5/3 and passes through the point (–2, 7). What

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Math Expert
Joined: 02 Sep 2009
Posts: 52971
Line A has a slope of -5/3 and passes through the point (–2, 7). What  [#permalink]

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15 Mar 2015, 22:04
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79% (01:55) correct 21% (02:20) wrong based on 249 sessions

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Line A has a slope of -5/3 and passes through the point (–2, 7). What is the x-intercept of Line A?

A. 2 1/3
B. 2 1/5
C. 2 2/3
D. 2 2/5
E. 2 3/5

Kudos for a correct solution.

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Re: Line A has a slope of -5/3 and passes through the point (–2, 7). What  [#permalink]

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16 Mar 2015, 21:55
1
2
Hi All,

Since it looks like this Ds question has gotten 'buried', I'm going to bump it back up with a 'nudge' about how to solve it:

When dealing with graphing and lines, it helps to organize your data using slope-intercept format:

Y = (M)(X) + B

We're given the slope of a line and one of the points that it runs through. From this, you can calculate the value of B. With the equation of the full line, you should then be able to calculate the X-intercept....

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Line A has a slope of -5/3 and passes through the point (–2, 7). What  [#permalink]

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16 Mar 2015, 23:04
1
2
slope -5/3 and point (-2;7), so equation is

7=(-5/3)*-2+b => b=11/3

to find X intercept, rewrite the equation

0=(-5/3)*x+11/3 => x=11/5=2 1/5

B
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Re: Line A has a slope of -5/3 and passes through the point (–2, 7). What  [#permalink]

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17 Mar 2015, 08:31
1
1
Here is my algebraic approach:
y = (-5/3)x + b
7 = (-5/3)(-2) + b
b = 11/3

0 = (-5/3)x + 11/3
x = 11/5 or 2 1/5
Math Expert
Joined: 02 Sep 2009
Posts: 52971
Line A has a slope of -5/3 and passes through the point (–2, 7). What  [#permalink]

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23 Mar 2015, 02:58
Bunuel wrote:
Line A has a slope of -5/3 and passes through the point (–2, 7). What is the x-intercept of Line A?

A. 2 1/3
B. 2 1/5
C. 2 2/3
D. 2 2/5
E. 2 3/5

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

Think about this visually. A slope of – 5/3 means, among other things, right 3 spaces, down 5. If the line goes through the point (–2, 7), then it must also go through (1, 2), which is much closer to the x-intercept. Let’s think about this in the vicinity of the x-intercept.

Obviously, (1, 2) is at a height of 2 above the x-axis. Let b be the distance from (1, 0) to the x-intercept. We know –h/b must equal the slope.

h/b = 5/3 --> 2/b = 5/3.
5b = 6 --> b = 6/5 = 1 1/5.

Now, we just have to add one to that to get the horizontal distance from the origin.

x-intercept = 2 1/5.

Attachment:

ppocg_img6.png [ 9.41 KiB | Viewed 5698 times ]

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Line A has a slope of -5/3 and passes through the point (–2, 7). What  [#permalink]

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25 May 2015, 04:05
1
Bunuel wrote:
Bunuel wrote:
Line A has a slope of -5/3 and passes through the point (–2, 7). What is the x-intercept of Line A?

A. 2 1/3
B. 2 1/5
C. 2 2/3
D. 2 2/5
E. 2 3/5

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

Think about this visually. A slope of – 5/3 means, among other things, left 3 spaces, down 5. If the line goes through the point (–2, 7), then it must also go through (1, 2), which is must closer to the x-intercept. Let’s think about this in the vicinity of the x-intercept.
Attachment:
ppocg_img6.png

Obviously, (1, 2) is at a height of 2 above the x-axis. Let b be the distance from (1, 0) to the x-intercept. We know –h/b must equal the slope.

h/b = 5/3 --> 2/b = 5/3.
5b = 6 --> b = 6/5 = 1 1/5.

Now, we just have to add one to that to get the horizontal distance from the origin.

x-intercept = 2 1/5.

Hi Bunuel,

Please make me understand the below point:

A slope of – 5/3 means, among other things, left 3 spaces, down 5. If the line goes through the point (–2, 7), then it must also go through (1, 2), which is must closer to the x-intercept.

I believe, for slope - 5/3, right 3 spaces and down 5 so that the line goes through (1,2) from (-2,7) downward slope. Please explain if iam wrong.

Thanks
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Joined: 02 Sep 2009
Posts: 52971
Re: Line A has a slope of -5/3 and passes through the point (–2, 7). What  [#permalink]

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25 May 2015, 07:54
balamoon wrote:
Bunuel wrote:
Bunuel wrote:
Line A has a slope of -5/3 and passes through the point (–2, 7). What is the x-intercept of Line A?

A. 2 1/3
B. 2 1/5
C. 2 2/3
D. 2 2/5
E. 2 3/5

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

Think about this visually. A slope of – 5/3 means, among other things, left 3 spaces, down 5. If the line goes through the point (–2, 7), then it must also go through (1, 2), which is must closer to the x-intercept. Let’s think about this in the vicinity of the x-intercept.

Obviously, (1, 2) is at a height of 2 above the x-axis. Let b be the distance from (1, 0) to the x-intercept. We know –h/b must equal the slope.

h/b = 5/3 --> 2/b = 5/3.
5b = 6 --> b = 6/5 = 1 1/5.

Now, we just have to add one to that to get the horizontal distance from the origin.

x-intercept = 2 1/5.

Hi Bunuel,

Please make me understand the below point:

A slope of – 5/3 means, among other things, left 3 spaces, down 5. If the line goes through the point (–2, 7), then it must also go through (1, 2), which is must closer to the x-intercept.

I believe, for slope - 5/3, right 3 spaces and down 5 so that the line goes through (1,2) from (-2,7) downward slope. Please explain if iam wrong.

Thanks

Yes, that's true. Edited. Thank you.
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Re: Line A has a slope of -5/3 and passes through the point (–2, 7). What  [#permalink]

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26 May 2015, 00:55
1
1
My attempt

I am very fond of co-ordinate geometry

Using slope intercept form of equation y=mx+c
In this case
y=(-5/3)x+c

We know that the line passes through (-2,7)
So solving for c
c=7-(5*2/3)=11/3

So equation of line is
y=(-5/3)x+(11/3)

Changing the equation of line to intercept form
x/(11/5) + y/(11/3) = 1

So x intercept is 11/5=2 1/5 which is B

Some theory to keep in mind

An equation of line can be represented in multiple forms
1. y=mx+c where m is the slope and c is a constant
2. x/a + y/b = 1 here a and b are the intercepts on x and y axis respectively
3. (y-y1)/(y2-y1) = (x-x1)/(x2-x1) equation of line passing through (x1,y1) and (x2,y2)

I think basis these it becomes easier to solve and one must know how to change forms of these equations quickly. All these are linear equations.
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Posts: 21
Re: Line A has a slope of -5/3 through the point (–2, 7). What  [#permalink]

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26 May 2015, 04:26
With the equation y=mx+c we get
7=10/3 + c
So c = 11/3

To find the x intercept put y=0 in the equationy=mx+c

0= -5/3 x +11/3
Multiplying by 3 throughout
-11 = -5x
So x=11/5
Which is 2 1/5
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Joined: 01 Apr 2014
Posts: 4
Re: Line A has a slope of -5/3 and passes through the point (–2, 7). What  [#permalink]

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07 Apr 2016, 22:57
Jackal wrote:
My attempt

I am very fond of co-ordinate geometry

Using slope intercept form of equation y=mx+c
In this case
y=(-5/3)x+c

We know that the line passes through (-2,7)
So solving for c
c=7-(5*2/3)=11/3

So equation of line is
y=(-5/3)x+(11/3)

Changing the equation of line to intercept form
x/(11/5) + y/(11/3) = 1

So x intercept is 11/5=2 1/5 which is B

Some theory to keep in mind

An equation of line can be represented in multiple forms
1. y=mx+c where m is the slope and c is a constant
2. x/a + y/b = 1 here a and b are the intercepts on x and y axis respectively
3. (y-y1)/(y2-y1) = (x-x1)/(x2-x1) equation of line passing through (x1,y1) and (x2,y2)

I think basis these it becomes easier to solve and one must know how to change forms of these equations quickly. All these are linear equations.

HOW YOU ARE EQUATING 11/5 TO 21/5 PLEASE EXPLAIN
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Joined: 02 Aug 2009
Posts: 7334
Re: Line A has a slope of -5/3 and passes through the point (–2, 7). What  [#permalink]

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07 Apr 2016, 23:04
akshay4gmat wrote:
HOW YOU ARE EQUATING 11/5 TO 21/5 PLEASE EXPLAIN

Hi here 11/5 is not 21/5 BUT $$2 \frac{1}{5}$$that is 2 and 1/5..
$$2 + \frac{1}{5} = \frac{(10+1)}{5} =\frac{11}{5}$$
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Re: Line A has a slope of -5/3 and passes through the point (–2, 7). What  [#permalink]

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08 Apr 2016, 06:46
akshay4gmat wrote:
Jackal wrote:
My attempt

I am very fond of co-ordinate geometry

Using slope intercept form of equation y=mx+c
In this case
y=(-5/3)x+c

We know that the line passes through (-2,7)
So solving for c
c=7-(5*2/3)=11/3

So equation of line is
y=(-5/3)x+(11/3)

Changing the equation of line to intercept form
x/(11/5) + y/(11/3) = 1

So x intercept is 11/5=2 1/5 which is B

Some theory to keep in mind

An equation of line can be represented in multiple forms
1. y=mx+c where m is the slope and c is a constant
2. x/a + y/b = 1 here a and b are the intercepts on x and y axis respectively
3. (y-y1)/(y2-y1) = (x-x1)/(x2-x1) equation of line passing through (x1,y1) and (x2,y2)

I think basis these it becomes easier to solve and one must know how to change forms of these equations quickly. All these are linear equations.

HOW YOU ARE EQUATING 11/5 TO 21/5 PLEASE EXPLAIN

Hi Akshay

Just as Chetan explained.

$$= \frac{11}{5}$$

$$= \frac{(10 + 1)}{5}$$

$$= 2 \frac{1}{5}$$

Apologies for not using Latex equivalent symbols
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Re: Line A has a slope of -5/3 and passes through the point (–2, 7). What  [#permalink]

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14 Jul 2016, 06:55
1
Bunuel wrote:
Line A has a slope of -5/3 and passes through the point (–2, 7). What is the x-intercept of Line A?

A. 2 1/3
B. 2 1/5
C. 2 2/3
D. 2 2/5
E. 2 3/5

Kudos for a correct solution.

Point intercept form a line is y=mx+b
slope = m,
y intercept =b (given that x=0)
x intercept =-b/m
Here we know m=-5/3 ; we need to find b and we are done

lets use the equation by plugging values of x,y and m from the question stem
7=-5/3*-2 +b ==> 7=10/3+b

b=11/3

x intercept = -b/m
= -11/3/-5/3=-11/-5==> 11/5
= 2 $$\frac{1}{5}$$

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Re: Line A has a slope of -5/3  [#permalink]

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03 Jun 2017, 23:39
slope m= -5/3 and point (-2;7), so equation is

y=mx+b

We need to find b
7=(-5/3)*-2+b
7-10/3=b

b=11/3
to find x intercept, rewrite the equation

0=(-5/3)*x+11/3 => x=11/5

Press +1 Kudos if you like my answer
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Joined: 08 Sep 2016
Posts: 32
Re: Line A has a slope of -5/3 and passes through the point (–2, 7). What  [#permalink]

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05 Nov 2017, 02:45
Hi @bunnel,

Could please explain the how you arrived at x intercept.

The equation of the line is -5x-3y+11 = 0

So should the y intercept be -a/b = -5/3?

S

Jackal wrote:
My attempt

I am very fond of co-ordinate geometry

Using slope intercept form of equation y=mx+c
In this case
y=(-5/3)x+c

We know that the line passes through (-2,7)
So solving for c
c=7-(5*2/3)=11/3

So equation of line is
y=(-5/3)x+(11/3)

Changing the equation of line to intercept form
x/(11/5) + y/(11/3) = 1

So x intercept is 11/5=2 1/5 which is B

Some theory to keep in mind

An equation of line can be represented in multiple forms
1. y=mx+c where m is the slope and c is a constant
2. x/a + y/b = 1 here a and b are the intercepts on x and y axis respectively
3. (y-y1)/(y2-y1) = (x-x1)/(x2-x1) equation of line passing through (x1,y1) and (x2,y2)

I think basis these it becomes easier to solve and one must know how to change forms of these equations quickly. All these are linear equations.
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Joined: 02 Sep 2016
Posts: 670
Re: Line A has a slope of -5/3 and passes through the point (–2, 7). What  [#permalink]

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10 Nov 2017, 02:46
Is this solution fine?

y= mx+c
y= -5/3*x +c

Using formula to find slope:
(7-y)/(-2-x) = -5/3

3(7-y)= -5(-2-x)

21-3y= 10+5x

5x+3y= 11

To find x-intercept, plug y= 0

x= 11/5
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Re: Line A has a slope of -5/3 and passes through the point (–2, 7). What  [#permalink]

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14 Nov 2017, 06:10
1
Bunuel wrote:
Line A has a slope of -5/3 and passes through the point (–2, 7). What is the x-intercept of Line A?

A. 2 1/3
B. 2 1/5
C. 2 2/3
D. 2 2/5
E. 2 3/5

The equation of Line A is y = (-5/3)x + b.

Let’s now plug in (-2,7) and determine b:

7 = (-5/3)(-2) + b

7 = 10/3 + b

Multiplying by 3, we have:

21 = 10 + 3b

11 = 3b

11/3 = b

Thus, we have:

y = (-5/3)x + 11/3

We can substitute 0 for y to determine the x-intercept, and we have:

0 = (-5/3)x + 11/3

-11/3 = -5x/3

-11 = -5x

x = -11/-5 = 11/5 = 2 1/5

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Line A has a slope of -5/3 and passes through the point (–2, 7). What  [#permalink]

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15 Nov 2017, 12:39
1
Bunuel wrote:
Line A has a slope of -5/3 and passes through the point (–2, 7). What is the x-intercept of Line A?

A. 2 1/3
B. 2 1/5
C. 2 2/3
D. 2 2/5
E. 2 3/5

Kudos for a correct solution.

With a slope and a point, and asked to find the x-intercept of a line, I use the slope intercept form. (It's quick).

1) Start with just the slope to construct a slope-intercept equation for the line:
$$y = mx + b$$, m = slope, b = y-intercept
Slope = $$-\frac{5}{3}$$,so
$$y = -\frac{5}{3}x + b$$

2) Next, plug in the given point (-2,7) to find $$b$$
$$7 = -\frac{5}{3}(-2) + b$$
$$7 = \frac{10}{3} + b$$
$$(7 - \frac{10}{3}) = \frac{11}{3} = b$$

3) Plug $$b$$ in. The line's equation is now
$$y = -\frac{5}{3}x + \frac{11}{3}$$

4) Set y equal to 0 to find x-intercept
$$0 = -\frac{5}{3}x + \frac{11}{3}$$
$$\frac{5}{3}x=\frac{11}{3}$$
$$x=\frac{11}{3}*\frac{3}{5}=\frac{11}{5}$$
$$=2\frac{1}{5}$$

Shiv2016 , yes, it's correct. I had to grin, though (with relief!) because I have done what you did . . . There is no need to find the slope. It's given.

shinrai , you wrote
Quote:
Could please explain the how you arrived at x intercept?

The equation of the line is -5x-3y+11 = 0

So should the y intercept be -a/b = -5/3?

Mmm, no. The y-intercept is 11/3. -5/3 is the slope. You were quoting Jackal, who wrote, "So equation of line is y=(-5/3)x+(11/3)"

Jackal got the x-intercept by setting y equal to 0:
0 = (-5/3)x+(11/3)
$$\frac{5}{3}x = \frac{11}{3}$$
$$x=\frac{11}{3} * \frac{3}{5}=\frac{11}{5}$$

The line crosses the x-axis at $$\frac{11}{5}$$ (and when it crosses the x-axis, y = 0)

You used this equation for the line: -5x-3y+11 = 0. (A more standard form is 5x + 3y = 11.)

It doesn't matter which form of equation you use; to find the y-intercept, set x = 0. (x IS zero when the line crosses the y-axis.) Your way:

-5x - 3y + 11 = 0
(-5)(0) -(3)(y) + 11 = 0
-3y = -11
y = $$\frac{11}{3}$$

Hope it helps.
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Line A has a slope of -5/3 and passes through the point (–2, 7). What   [#permalink] 15 Nov 2017, 12:39
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