Question

Find the distance of the point (3,2) from the straight line whose slope is 5 and is passing through the point of intersection of lines x + 2y = 5 and x – 3y + 5 = 0

Answer 1

by solving the two equation simultaneously
you'll get point of intersection as (1,2)
Now the req eq of st line whose slope is 5 is
y-2=5 (x-1)
i.e.
5x-y-3=0
perpendicular distanxe from the point to the line is
$\frac{5x - y - 3}{ \sqrt{ {5}^{2} } + {1}^{2} }$
put the values of x and y and calculate the distance

Answer 2

Answer:

Step-by-step explanation:

by solving the two equation simultaneously

you'll get point of intersection as (1,2)

Now the req eq of st line whose slope is 5 is

y-2=5 (x-1)

i.e.

5x-y-3=0

perpendicular distanxe from the point to the line is

put the values of x and y and calculate the distance