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avatar+154 

In the standard (x,y) coordinate plane, a line passes through the points (1,-2) and (5,10). At which of the following points does the line cross the y-axis?

 

a. (-8,0)

b. (-5,0)

c. (0,0)

d. (0,-8)

e. (0,-5)

ISmellGood  Aug 24, 2017
edited by ISmellGood  Aug 24, 2017
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4+0 Answers

 #1
avatar+76108 
+2

 

(1, - 2)   and  (5,10).....we can find the slope  of a line connecttng these points thusly :

 

[  10 - - 2 ] / [ 5 - 1 ]  =  12 / 4  = 3

 

The equation of the said line is :

 

y  = 3 ( x - 5)  + 10

 

y = 3x - 15 + 10

 

y = 3x  - 5        where this line crosses the y axis, x  = 0....so

 

y = 3(0)  - 5   =  0 - 5  =  -5

 

So.....the line crosses the y axis at  ( 0, - 5)

 

Here's a graph : https://www.desmos.com/calculator/eearzhfvuk

 

cool cool cool

CPhill  Aug 24, 2017
 #3
avatar+1828 
0

Is the answer a, b, c, or d?

gibsonj338  Aug 24, 2017
 #4
avatar+154 
0

Thanx, I forgot to add the last option haha

ISmellGood  Aug 24, 2017
 #2
avatar+1828 
0

To answer the question, first figure out the equaton of the line. To do that first find the slope.  The formula for the slope of a line is

 

\(m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\) where m = slope, \({y}_{2}\) = y-coordinate in the second point, \({y}_{1}\) = y-coordinate in the first point, \({x}_{2}\) = x-coordinate in the second point, and \({x}_{1}\) = x-coordintate in the first point.

 

\(m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\)

 

m = ?

 

\({y}_{2}\) = 10

 

\({y}_{1}\) = -2

 

\({x}_{2}\) = 5

 

\({x}_{1}\) = 1

 

\(m=\frac{10-(-2)}{5-1}\)

 

\(m=\frac{10+2}{5-1}\)

 

\(m=\frac{12}{5-1}\)

 

\(m=\frac{12}{4}\)

 

\(m=\frac{3}{1}\)

 

\(m=3\)

 

Now that we know that the slope of the line is 3, put that in the equation for a line.  The equation for a line is

 

\(y=mx+b\) where \(y\) = y-coordinate, \(m\) = slope, \(x\) = x-coordinate, and \(b\) = y intercept (where line crosses the y axis).  Take one of the points and susitute \(x\) and \(y\) in the equation so we can solve for b.

 

\(y=mx+b\)

 

\(y\) = 10

 

\(m\) = 3

 

\(x\) = 5

 

\(b\) = ?

 

\(10=3\times5+b\)

 

\(10=15+b\)

 

\(10-15=15+b-15\)

 

\(-5=15+b-15\)

 

\(-5=b-0\)

 

\(-5=b\)

 

\(b=-5\)

 

Now fill in what you know leaving \(y\) and \(x\) as \(y\) and \(x\).

 

\(y=3x-5\)

 

To figure out at which point the line crosses the y-axis, subsitute \(x\) for 0 and solve for \(x\).

 

\(y=3x-5\)

 

\(y=3\times0-5\)

 

\(y=0-5\)

 

\(y=-5\)

 

The point that the line crosses the y-axis is at point \((0,-5)\) which means that the answer is neither a, b, c, or d.

gibsonj338  Aug 24, 2017

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