Question

find the equation of a straight line perpendicular to the line y=4/3x-7 and passing the point (7,-1)​

Answer 1

Answer:

x+12y-6=0

Step-by-step explanation:

GT-

H(7,-1)

y=4(3x-7)

y=12x-28

Slope,m=12

Since the lines are perpendicular

Therefore,

m=-1/12

Required equ.:

y+1=-1/12(x-7)

12y+1=-x+7

x+12y-6=0

HOPE THIS HELPS............

Answer 2

Given,

Equation of line y = 4/3x -7 and passing through the point (7,-1).

To find,

Find the equation of a straight line that is perpendicular to the line y=4/3x-7 and passing through the point (7,-1)​.

Solution,

• Since the product of slopes of the line perpendicular to each other is -1.

        Mathematically it can be represented as m1×m2 = -1.

• The slope of line y = 4/3x -7 is 4/3 so, the slope of the line perpendicular to the equation y = 4/3x -7 should be -3/4.

• The perpendicular line is passing through (7,-1).

        Hence the equation of line is (y -(-1)) / (x - 7) = -3/4.

⇒   4(y+1) = -3(x-7).

⇒   4y + 4 = -3x + 21

⇒   4y + 3x + 4 - 21 = 0

⇒   3x + 4y - 17 = 0.

Hence, the equation of a straight line perpendicular to the line y=4/3x-7 and passing through the point (7,-1)​ is 3x + 4y - 17 = 0.