a) x – 2y + 6 = 0 b) x + y = 1 c) y = 1 d) y = -3x + 8
<h3><u>Solution:</u></h3>
<em><u>a. The points (-4,1) and (2,4) both lie on the line</u></em>
The general line equation on which (a, b) and (c, d) lies is:
![y-\mathrm{b}=\frac{d-b}{c-a}(x - a)](https://tex.z-dn.net/?f=y-%5Cmathrm%7Bb%7D%3D%5Cfrac%7Bd-b%7D%7Bc-a%7D%28x%20-%20a%29)
Here the given points are (a, b) = (-4, 1) and (c, d) = (2, 4)
Thus the required equation is:
![y-1=\frac{4-1}{2-(-4)}(x-(-4))](https://tex.z-dn.net/?f=y-1%3D%5Cfrac%7B4-1%7D%7B2-%28-4%29%7D%28x-%28-4%29%29)
On solving we get,
![\begin{array}{l}{\rightarrow y-1=\frac{3}{2+4}(x+4)} \\\\ {\rightarrow y-1=\frac{3}{6}(x+4)} \\\\ {\rightarrow 2(y-1)=1(x+4)} \\\\ {\rightarrow 2 y-2=x+4} \\\\ {\rightarrow x-2 y+6=0}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B%5Crightarrow%20y-1%3D%5Cfrac%7B3%7D%7B2%2B4%7D%28x%2B4%29%7D%20%5C%5C%5C%5C%20%7B%5Crightarrow%20y-1%3D%5Cfrac%7B3%7D%7B6%7D%28x%2B4%29%7D%20%5C%5C%5C%5C%20%7B%5Crightarrow%202%28y-1%29%3D1%28x%2B4%29%7D%20%5C%5C%5C%5C%20%7B%5Crightarrow%202%20y-2%3Dx%2B4%7D%20%5C%5C%5C%5C%20%7B%5Crightarrow%20x-2%20y%2B6%3D0%7D%5Cend%7Barray%7D)
<em><u>b.) m= -1 and the point (2, -1) lies on the line</u></em>
The equation of line in point slope form is y – b = m(x – a)
where m is slope and (a, b) is a point on it
Here m = -1 and (a, b) = (2, -1)
Thus the required equation is:
y – (-1) = -1(x - 2)
y + 1 = -x + 2
y = -x + 2 -1
y = -x + 1
<em><u>c. )It has the same slope as y = 5 and passes through (1, 1)</u></em>
our line has same slope with y = 5, then our equation would be y = k and it passes through (x, y) = (1, 1) so, then by substitution
1 = k
k =1
Then our equation will be y = k
y = 1
<em><u>
d. ) m= -3 and it has a y-intercept of (0, 8)</u></em>
line equation in slope intercept form is y = mx + b where m is slope and b is y – intercept.
Then, our equation will be y = -3x + 8
We took y- intercept = 8 as it is the value of y when x = 0