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Question: How do you determine the linear function whose graph is a line that contains the points \(\left( { -...

How do you determine the linear function whose graph is a line that contains the points (1,8)\left( { - 1, - 8} \right) and (2,10)\left( {2,10} \right) ?

Explanation

Solution

The equation of a line that passes through two points can be calculated by using the two-point form of the equation of a line. We will substitute the given two points into the two-point form of a line. Then after simplifying we can determine the required linear function.

Formula used:
The two-point form of a line passing through the points (x1,y1)\left( {{x_1},{y_1}} \right) and (x2,y2)\left( {{x_2},{y_2}} \right) is given by yy1xx1=y2y1x2x1\dfrac{{y - {y_1}}}{{x - {x_1}}} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} .

Complete step-by-step answer:
As we know we can determine the equation of a line that passes through two points by using the two-point form of a line.
Now, the two given points are
A=(1,8)A = \left( { - 1, - 8} \right) and B=(2,10)B = \left( {2,10} \right) .
Now, we know that the first coordinate is the x-coordinate and the second coordinate is the y-coordinate.
Here,
For point AA, x1=1{x_1} = - 1 and y1=8{y_1} = - 8
For point BB, x2=2{x_2} = 2 and y2=10{y_2} = 10
Now, as we know that, the two-point form of a line is
yy1xx1=y2y1x2x1\dfrac{{y - {y_1}}}{{x - {x_1}}} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} .
Now, we can determine the linear function by substituting the x1=1{x_1} = - 1 , x2=2{x_2} = 2 , y1=8{y_1} = - 8 and y2=10{y_2} = 10 in the two-point form of the equation of a line.
After substituting the values, we get
y(8)x(1)=10(8)2(1)\dfrac{{y - ( - 8)}}{{x - ( - 1)}} = \dfrac{{10 - ( - 8)}}{{2 - ( - 1)}}
Now, after opening the brackets of numerators as well as denominators on both sides, we get
y+8x+1=10+82+1\Rightarrow \dfrac{{y + 8}}{{x + 1}} = \dfrac{{10 + 8}}{{2 + 1}}
y+8x+1=183\Rightarrow \dfrac{{y + 8}}{{x + 1}} = \dfrac{{18}}{3}
On Right-hand side, simplifying by dividing 18  18\; by 33 , we have
y+8x+1=61\Rightarrow \dfrac{{y + 8}}{{x + 1}} = \dfrac{6}{1}
Now, by cross multiplying, we get
1(y+8)=6(x+1)1(y + 8) = 6(x + 1)
y+8=6x+6\Rightarrow y + 8 = 6x + 6
Now, by subtracting 88 on both sides, we get
\Rightarrow y+88=6x+68y + 8 - 8 = 6x + 6 - 8
y=6x2\Rightarrow y = 6x - 2

y=6x2y = 6x - 2 is the required linear function whose graph is a line that passes through the points (1,8)( - 1, - 8) and (2,10)(2,10).

Note:
There is an alternate way to prove the two points form of the equation of a straight line. Consider the point-slope form of the equation of a line,
we have, yy1=m(xx1)y - {y_1} = m(x - {x_1}) - - - - - - (1.)(1.)
Since the line is passing through the point (x1,y1)\left( {{x_1},{y_1}} \right) in (1.)\left( {1.} \right) and the slope of the line is m=y2y1x2x1m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} . So, (1.)(1.) becomes yy1=(y2y1x2x1)(xx1)y - {y_1} = \left( {\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right)\left( {x - {x_1}} \right).