Question

Write the point-slope form of the equation of the line through point (6, -2) that is perpendicular to the line y=2x-3​

Answer 1

Answer:

The point - slope form of the equation of line 1 is y = (-1/2)x + 1

Step-by-step explanation:

here, the given point on the equation  1  is A(6, -2).

Now, equation of line 2 is y = 2x - 3

Comparing it with the INTERCEPT SLOPE FORM :  y =  mx + C

Slope of Line 2    = 2  (= m2)

Now,as line 1 is perpendicular to line 1

⇒ Slope of line 1  x Slope of line 2  = -1

or, slope of line 1  = (-1/2)

Now, by POINT SLOPE form of a equation:

An equation with point (x0 , y0) and slope m is given as

(y - y0)=  m (x - x0)

Here, the equation of line 1 with point (6, -2) and slope (-1/2) is given as:

$y - (-2) = \frac{-1}{2} (x -6)\\\implies 2( y + 2) = 6 -x\\or, 2y + 4 = 6 -x\\\implies 2y = -x + 2\\or, y = (-\frac{1}{2}) x + 1$

Hence, the point - slope form of the equation of line 1 is y = (-1/2)x + 1