Question

Find the equation of the straight line passing through (3,- 5) and parallel to the
line joining the points (1, 2) and (-3, 4).​

Answer 1

Step-by-step explanation:

As the required line is parallel to line formed by (1, 2) & (- 3, 4),  slope of lines must be same.

Slope of required line = slope of line by (1,2) & (-3, 4)

                    = (4 - 2)/(- 3 - 1)

                    = - 1/2

Hence the required line from (3, - 5) is;

⇒ y - (-5) = (-1/2)(x - 3)

⇒ y + 5 = (-1/2)(x - 3)

⇒ 2(y + 5) = - (x - 3)

⇒ 2y + 10 = - x + 3

⇒ x + 2y + 7 = 0  

Answer 2

Given ,

The straight line passing through the point (3,- 5) and parallel to the line joining the points (1,2) and (-3,4)

We know that , the point slope form is given by

$\boxed{ \tt{Slope \: (m) = \frac{ y_{2} -y_{1} }{x_{2} -x_{1}} }}$

If two lines are parallel to each other , then

$\boxed{ \tt{m_{1} = m_{2}}}$

Thus , the slope of line passing through points (1, 2) and (-3, 4) , will be

m = ( 4 - 2)/( - 3 - 1)

m = 2/(-4)

m = -1/2

Since , the two lines are // to each other

Therefore , the slope of line passing through (3 , -5) is -1/2

Now , the line whose slope is -1/2 is passing through (3,-5)

Thus , the equation of the line will be

-1/2 = (-5 - y)/(3 - x)

-3 + x = -10 - 2y

x + 2y + 7 = 0

Therefore , the required slope of the line is x + 2y + 7 = 0

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