Question

Find the equation of the line that is perpendicular to y = −4x + 10 and passes though the point (7,2).

Answer 1

EXPLANATION.

Equation of the line perpendicular to : y = - 4x + 10.

Passes through the point = (7,2).

As we know that,

Slope of the perpendicular line = b/a.

Slope of the line = y = - 4x + 10.

⇒ y + 4x - 10 = 0.

Slope = b/a = 1/4. = M

As we know that,

Equation of the line,

⇒ (y - y₁) = m(x - x₁).

Put the values in the equation, we get.

⇒ (y - 2) = 1/4(x - 7).

⇒ 4(y - 2) = 1(x - 7).

⇒ 4y - 8 = x - 7.

⇒ 4y - x - 8 + 7 = 0.

⇒ 4y - x - 1 = 0.

                                                                                                                                             

MORE INFORMATION.

NOTE.

(1) = Equation of a line which is parallel to ax + by + c = 0 is ax + by + k = 0.

(2) = Equation of a line which is perpendicular to ax + by + c = 0 is bx - ay + k = 0.

The value of k in both cases is obtained with the help of additional information given in the problems.

Answer 2

Answer:

Given :-

y = -4x + 10

Point (7,2)

To Find :-

Equation line

Solution :-

Here,

$\large \sf \: y + 4x - 10 = 0$

Slope = M = 1/4

Now,

$\large \sf \implies \: (y - 2) = \dfrac{1}{4} (x - 7)$

$\large \sf \implies \: 4(y - 2) = 1(x - 7)$

$\large \sf \implies \: 4y - 8 = x - 7$

$\large \sf \implies \: 4y - x - (8 - 7) = 0$

$\mathfrak \red{4y - x - 1 = 0}$