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

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

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.

Answer = (B)

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.

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.

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

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..
\(2 + \frac{1}{5} = \frac{(10+1)}{5} =\frac{11}{5}\)

Hi Akshay

Just as Chetan explained.

\(= \frac{11}{5}\)

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

\(= 2 \frac{1}{5}\)

Apologies for not using Latex equivalent symbols

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 \(\frac{1}{5}\)

Answer is B

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

Answer B

Press +1 Kudos if you like my answer

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?

Thanks in advance!
S

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

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

Answer: B

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
\(0 = -\frac{5}{3}x + \frac{11}{3}\)
\(\frac{5}{3}x=\frac{11}{3}\)
\(x=\frac{11}{3}*\frac{3}{5}=\frac{11}{5}\)
\(=2\frac{1}{5}\)

Answer B

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

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.

if someone is wondering why the second equation is equal to zero- it is because while calculating X intercept, Y will be 0.

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