Find the equation of the line which goes through the point (3,10) and is parallel to the line 7x-y=1.
Answers
Answer:
y - 7x - 11 = 0
Step-by-step explanation:
Given that a line passes through the point (3,10) and is parallel to the line 7x - y = 1. If a line is parallel to another line, this implies that slope of both the lines is same i.e. steepness of both lines is same.
Slope of a line is given by, -A/B where A is coefficient of x and B is coefficient of y.
Slope of line 7x - y = 1 is given by, -7/-1 = 7.
Slope of the required line will also be equal to 7.
Now we have the information that a line whose slope is 7 and passes through the point (3,10) and we have to find it's equation.
Whenever we are given slope and one point from which a line passes, we use the slope point form of straight line which is given by,
- y - y1 = ( x - x1 ) m
Here,
- x1 and y1 are the points from which the given line passes.
- m is the slope of the line.
By substituting value of x1, y1 and m, we get :
=> y - 10 = (x - 3)7
=> y - 10 = 7x - 21
=> y - 7x - 10 + 21 = 0
=> y - 7x - 11 = 0
This is the required equation of straight line.