Question

Find the equation of a line passing through the points (-3, 1) and (1, 5)

x-y=8
x-y+4=0
x+y=4

Answer 1

To find point of intersection

solve equation of both lines

2x+y=5 …………(1)

x+3y+8=0 …………..(2)

Point of inter-section=(

5

23

,

5

−21

)

The required line is parallel to 3x+4y=7

Line parallel to ax+by+c

1

=0 is ax+by+λ=0 required line is 3x+4y+

i

^

=0

We know that it passes through (

5

23

,

5

−21

)

=3(

5

23

)−4(

5

21

)+λ=0

=−(

5

69

−

5

84

)=λ

λ=(−3)

⇒λ

⇒3x+4y+3=0.

Answer 2

Step-by-step explanation:

$equation \: \: of \: \: line \: \: \\ \\ (y - y1) = ( \frac{y2 - y1}{x2 - x1} )(x - x1)$

(x1 , y1)=(-3,1)

(x2 , y2) = (1, 5)

$(y - 1) = \frac{4}{4} (x - ( - 3)) \\ \\ y - 1 = x + 3 \\ \\ x - y + 4 = 0$