Question

Find the equation of the line perpendicular to the line y = 3x-5 and passing through the point (-5, 8).

(help meeee)​

Answer 1

Equation of a line : y = 3x - 5

• As we know that,

General equation of a line : y = mx + c

where m is slope of that line, x is a variable and c is a constant term.

So, let's compare y = 3x - 5 with y = mx + c

• m = 3 = slope of y = 3x - 5
• m1 = 3

We also know that the product of perpendicular line is - 1 i.e m1×m2 = - 1

• Let's find another slope of a line

→ m1 × m2 = -1

→ m2 = -1/m1

→ m2 = -1/3

Coordinate of point of another line where it is passing i.e (-5,8)

• Let's find the equation of another line

→ y - y1 = m2(x - x1)

• m2 = -1/3
• x = -5
• y = 8

→ y - 8 = -1/3[x - (-5)]

→ y - 8 = -1/3(x + 5)

→ y - 8 = -x/3 - 5/3

→ y = -x/3 - 5/3 + 8

→ y = - x - 5 + 24/3

→ y = -x/3 + 19/3

Hence, required eqⁿ is y = -x/3 + 19/3

Answer 2

Answer:

Given:

• Find the equation of the line perpendicular to the line y = 3x-5 and passing through the point (-5, 8).

To Find:

• Find the equation of the line perpendicular
• y = mx + c

Solution:

As we know ,

• General equation of this line y = mx + c

Now, let's compare y = 3x - 5 with y = mx + c

• m = 3 = slope of y = 3x - 5
• m1 = 3

⤵️Let's find slope of line:

➢ m1 × m2 = -1

➢ m2 = -1/3

⤵️Let's find equation of another line:

➢ y - y1 = m2 (x - x1)

• ⇝ m2 = -1/3
• ⇝ x = -5
• ⇝ y = 8

➢ y - 8 = -1/3 [ x - ( -5) ]

➢ y - 8 = -1/3 ( x + 5 )

➢ y = -x/3 - 5/3 + 8

➢ y = -x - 5 + 24/3

➢ y = -x/3 + 19/3

⏩Hence,

Equation is y = -x/3 + 19/3