Question

Find the equation of the straight line which is passing through the points (1, - 1 )and (3, 5)​

Answer 1

Answer:

$x1 = 1 \: \: \: \: \: y1 = - 1 \: \: \: x2 = 3 \: \: y2 = 5 \\ slope(m) = \frac{y2 - y1}{x2 - x1} = \frac{6}{2} = 3 \\ equation \: of \: line \: = y - y1 = m(x - x1) \\ y + 1 = 3(x - 1) \\ y = 3x - 4 \\ \alpha \beta \gamma follow \: me$

Answer 2

We can use the equation of the line in two point form to find the equation of this line that is passing through the points (1,-1) and (3,5).

Let (x1,y1)=(1,−1)(x1,y1)=(1,−1) and (x2,y2)=(3,5)(x2,y2)=(3,5)

Equation of line in two point form is given as follows,

x−x1/x1−x2=y−y1/y1−y2

x−x1/x1−x2=y−y1/y1−y2

Substituting the values, we get,

x−11−3=y−(−1)−1−5x−11−3=y−(−1)−1−5

∴x−1−2=y+1−6∴x−1−2=y+1−6

∴x−11=y+13∴x−11=y+13

∴3(x−1)=y+1∴3(x−1)=y+1

∴3x−3−y−1=0∴3x−3−y−1=0

∴3x−y−4=0∴3x−y−4=0

This is the equation for the line passing through the points (1,-1) and (3,5). We can verify it by substituting both these points to the equation.

For (1,-1)

LHS=3(1)−(−1)−4LHS=3(1)−(−1)−4

=3+1−4=0=RHS=3+1−4=0=RHS

For (3,5)

LHS=3(3)−5−4=9−9=0=RHSLHS=3(3)−5−4=9−9=0=RHS

Hence, the equation is 3x−y−4=03x−y−4=0