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Question: Find the equation of straight line passing through the points \[( - 1,1)\] and \[(2, - 4)\]...

Find the equation of straight line passing through the points (−1,1)( - 1,1) and (2,−4)(2, - 4)

Explanation

Solution

Hint : Since two coordinate points are given and we want to find equation of straight line so we can directly use two point formula to find the equation of straight line and the formula is given by
y−y1x−x1=y2−y1x2−x1\dfrac{{y - {y_1}}}{{x - {x_1}}} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}

Complete step-by-step answer :
We know that equation of line through two points (x1,y1)({x_1},{y_1}) and (x2,y2)({x_2},{y_2})is
y−y1=y2−y1x2−x1(x−x1)y - {y_1} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}(x - {x_1})
Since the given equation of the line passes through the points (−1,1)( - 1,1) and (2,−4)(2, - 4).
Here, let us take
x1=−1,y1=1{x_1} = - 1,{y_1} = 1
x2=2,y2=−4{x_2} = 2,{y_2} = - 4
Substituting the values in the formula
(y−1)=−4−12−(−1)(x−(−1))(y - 1) = \dfrac{{ - 4 - 1}}{{2 - ( - 1)}}(x - ( - 1))
(y−1)=−52+1(x+1)(y - 1) = \dfrac{{ - 5}}{{2 + 1}}(x + 1)
(y−1)=−53(x+1)(y - 1) = \dfrac{{ - 5}}{3}(x + 1)
(cross multiply with 3)
3(y−1)=−5(x+1)3(y - 1) = - 5(x + 1)
after simplification we get
3y−3=−5x−53y - 3 = - 5x - 5
rearranging the terms we get-
5x+3y+2=05x + 3y + 2 = 0
Hence, the required equation is 5x+3y+2=05x + 3y + 2 = 0
So, the correct answer is “ 5x+3y+2=05x + 3y + 2 = 0”.

Note : we can solve this by using slope intercept form given by y=mx+cy = mx + c, where m is the slope of the given line and c is the y-intercept
First find the slope of the line using the formula
m = $$$$\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}
let us take
x1=−1,y1=1{x_1} = - 1,{y_1} = 1
x2=2,y2=−4{x_2} = 2,{y_2} = - 4
Substituting the values in the formula we get
m=m = −4−12−(−1)\dfrac{{ - 4 - 1}}{{2 - ( - 1)}}
now take any one given coordinate point as x and y that is x=−1,y=1x = - 1,y = 1 now substituting the values in the above equation we get
⇒1=−53(−1)+c\Rightarrow 1 = \dfrac{{ - 5}}{3}( - 1) + c
after simplification we get
⇒1=53+c\Rightarrow 1 = \dfrac{5}{3} + c
(shift 53\dfrac{5}{3} to left hand side we get)
⇒1−53=c\Rightarrow 1 - \dfrac{5}{3} = c
(taking lcm ,on simplifying we get)
⇒−23=c\Rightarrow \dfrac{{ - 2}}{3} = c
Now substitute c in equation 1 we get
⇒y=−53x+−23\Rightarrow y = \dfrac{{ - 5}}{3}x + \dfrac{{ - 2}}{3}
⇒y=−5x−23\Rightarrow y = \dfrac{{ - 5x - 2}}{3}
⇒3y=−5x−2\Rightarrow 3y = - 5x - 2 (shifting the terms and rearrange)
⇒5x+3y+2=0\Rightarrow 5x + 3y + 2 = 0
Hence, the required equation is 5x+3y+2=05x + 3y + 2 = 0
And it is same as solution obtained above